OLD VERSION: Probability - click the link in the description for an updated video
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- เผยแพร่เมื่อ 20 ต.ค. 2024
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• GMAT Ninja Quant Ep 15...
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thank you for such an optimistic approach to probability but unfortunately, I did spend a sleepless night on this and it was worth it. Seeing you effortlessly tutor us, my childhood dream of at least someone providing insightful knowledge instead of business in the name of education is fulfilled. thank you from the bottom of my heart😭😭😭
Once I get my desired score, the entire Ninja team will have a long thank-you speech from my side.
Thank you so much for the absolutely delightful comment! I'm sad that we cost you sleep with this video, but I'm honored that the video helped so much.
Have a great time studying, and please let us know how things go for you in the end! ❤️
For question 7, I think we could also evaluate (1) as the following:
When picking 2 dogs, there are 3 possibilities:
1. Get 2 Labradors
2. Get 1 Labrador and 1 non-Labrador
3. Get 0 Labradors
1+2+3 should add up to 100%. Statement (1) tells us the probability of picking 1 and 1 is >0.5; this implies prob 2 labradors + prob 0 labradros < 0.5, which in turn implies prob 2 labradors 0.
Haha, you're spot-on, Javier! And I'm laughing pretty hard at the fact that three of us collaborated on the content for this video, and none of us noticed that solution path IN A QUESTION THAT WE WROTE. It's funny: sometimes, we write questions with a certain method in mind, and that blinds us to the better solution path, even when it's staring us in the face.
Congrats on doing a better job with our question than I did. You deserve a cookie. :)
- Charles
@@GMATNinjaTutoringBut how does the same concept work on statement 2?
@@10aman I don't think it does work on statement 2. Unless I'm missing something (again!), there's no shortcut on statement 2, and it's probably easiest to do something similar to the solution shown in the video.
@@GMATNinjaTutoring I have a doubt. If x = 2 labs (2L), y= 1L+NL, z = 2NL, then since x+y+z = 1, if z > 0.1, then x can be any value between 0.9 and 0, including < 0.5 and >= 0.5 ; therefore, is it not not-sufficient? @GMATNinjaTutoring
@@GMATNinjaTutoring The same logic should work on st.2, right? prob (0 labras)>0.1 which means prob(1 labra)+prob(2 labras)
In Q4, how can you say number of male chefs > 25, If out of the 75 professionals you don't know how many are chefs/bartenders and how many are male/female ?
From statement (2), we know that there are 20 chefs. We don't need to know how many of these 20 chefs are male or female because even if every single one of these chefs was male, there would still be less than 25 male chefs at the conference.
Sometimes we don't need to know everything about a situation to answer the question. It depends on what question was asked and what information we were given. In this case, statement (2) is sufficient to answer the question even though we can't say for certain how many of the chefs and bartenders are male or female.
I hope that helps!
I absolutely love these videos - thank you so much for producing all this insightful content! Even if I don't get the score I am looking for, I feel so much more comfortable with the content of the GMAT given all these examples and tips. Very appreciated!
Thank you so much for the kind words, Megha! I'm glad that we've been able to help a bit. Good luck with your studies, and please let us know how things go for you!
Why is there no difference between right and left shoes in the first question?
Good question. The right-left thing implicitly baked into the solution, in a way.
In order to get a pair of pink shoes, your first selection needs to be ANY pink shoe -- during that first selection, you don't care whether it's the left or right shoe. You just need a pink shoe, and the probability of selecting one is 4/12 or 1/3, as shown in the video.
For the second selection, you want to select a pink shoe, but not just any pink shoe -- you want the one that matches your first selection. So if that first shoe was, say, a left pink Ugg, then you'll need the right pink Ugg. Similarly, if it was the right pink Ugg, you'll need the left.
Either way: exactly one shoe will match your first selection, regardless of whether you chose a left or right shoe on the first pick. So the probability of getting what you want on the second selection is 1/11. And 1/3 * 1/11 = 1/33, which gives us our final answer of (D).
I hope that helps!
The questions get tougher whenever Charles comes .....
Correct! When you see me coming, run. 😆
Have fun studying, Aditya!
Excellent! Aadaraniya Shri Charles Bibiliosji is one of the best GMAT instructors ever!
I'm honored, thank you so much for this, Ashish! You're making me blush. Have fun studying, and please keep us posted on your progress!
- Charles
@@GMATNinjaTutoring great class
In Q6, why do we multiply by 4 to include all the different arrangements? Is the question asking about arrangement? or just probability
In this question, we're asked what's the probability that exactly three of the four customers will purchase a pie. Let's say that we can call the four people A, B, C, and D to make things as simple as possible. Next, let's consider the scenario where A is the person who does not buy a pie, and B, C, and D all buy pies. To find the probability, we'd multiply (1/4)*(3/4)*(3/4)*(3/4) = 27/256
The reason we have to multiply by 4 is that we have to think about all the possible scenarios where one person does not buy a pie and the remaining three people do. There's a scenario in which B does not buy a pie and the other three do, there's a scenario in which C does not buy a pie and the other three do, and there's a scenario in which D does not buy a pie and the other three do.
The probability we found in the first paragraph is the same as each of the probabilities of the other scenarios in which another person doesn't buy a pie but the remaining three people do. This means that we can multiply 27/256 by 4 to take into account the 4 possible scenarios in which one person does not buy a pie and three people do. This gives us the answer of 27/64.
I hope that helps!
why cant i use the techinique used in the shoe pair question for the last one ten the answer would be (6/10)(5/9)(4/8)(3/7) = 1/14
You can use this method to answer the final question, but there are a few more steps to follow to reach the final answer and it gets quite complicated. The answer you've given suggests you pick the cookies in the following order: chocolate chip #1, chocolate chip #2, oatmeal #1, oatmeal #2, but there's nothing in the question to say we need to pick the cookies in that order. We could pick the same four cookies in any order and we'd still satisfy the conditions laid out in the question. This means we need to multiply our answer by the number of ways of rearranging the cookies. Since there are four cookies, we can rearrange these cookies in 4! different ways so we multiply your answer by 4! = 24.
But there's more! The two chocolate chip cookies are interchangeable, so it doesn't matter whether we select chocolate chip #1 and then chocolate chip #2 or we select chocolate chip #2 and then chocolate chip #1. This means the number of ways we can rearrange the cookies decreases by a factor of two, so we don't have to multiply your answer by 24, we have to multiply it by 24/2 = 12.
We can say exactly the same thing about the oatmeal cookies. Since they're interchangeable, we need to decrease the number we multiply your answer by another factor of two, so we now need to multiply it by 12/2 = 6.
If we take your answer and multiply it by 6, we get 6/14 = 3/7 which is the answer we're looking for. This is a perfectly valid way to answer this question but as you can see, it gets complicated quite quickly. Keeping track of when you need to multiply and when you need to divide gets tough, which is why we think the method Charles demonstrated in the video is a better way of solving this question. However, if this method makes more sense to you and you prefer it, then you should use this method.
I hope that helps!
i can't understand why is 3/4 the probability of ordering a beer, and why do you multiply 1/4
The question tells us the "probability that a customer at a certain bar will purchase a beer is 0.75." Converting the decimal 0.75 into a fraction gives us 3/4.
There's a full explanation of why you should find the probability that none of the four customers orders a beer, which is why you'd multiply 1/4 by itself four times, in the video starting at 13:20.
I hope that helps but let me know if you have any further questions.
Hi, for question 6, I don't understand why we can cancel out the exponent of 4 to the denominator of 4 of the fraction 1/4 ? Thank you so much for helping!
It's because the four on the outside of the set of parentheses wasn't an exponent. You could think of it as 4*[(3/4)*(3/4)*(3/4)*(1/4)]. Then the four on the outside of the parentheses cancels with one of the fours in the denominators of the fractions to give: [(3)*(3/4)*(3/4)*(1/4)].
What we're saying at that point is [(3/4)*(3/4)*(3/4)*(1/4)] represents the probability of one of the rows of possible scenarios Charles wrote on the board. Since there are four possible scenarios, we need to multiply this probability by 4 to get the final value. This gives us 4*[(3/4)*(3/4)*(3/4)*(1/4)].
I hope that helps!
@@GMATNinjaTutoring Oh I see, thanks! I was thinking it was an exponent and thought I was missing an important new rule of some sort haha. Thanks for the precision! Really appreciate!
In Question 1, Harry has 6 pairs of shoes. and out of that 2 are pink. shouldn't it be 4/12 * 2/11. 2 because If he chooses right in the first time, he is left with only 2 left side pink shoes.
It wouldn't be enough to just pick any pink left shoe to go with the right shoe he picked first. The question asks us to find the probability Harry chooses a "matching pink pair of shoes," so we're looking for the pink left shoe that completes the pair with the first shoe Harry chooses. There is only one shoe out of the remaining eleven that would make a matching pair, so the probability for the second shoe is 1/11.
I hope that helps!
@@GMATNinjaTutoring yeah thanks, it helps. I also just noticed that, question says he has different pairs of shoes.
Your videos are really amazing, thanks for these wonderful videos
Thanks for another awesome video Charles. Question on the last one: why does selecting two chocolate cookies translate to 6C2?
We'd like to choose two chocolate chip cookies from a group of six available cookies, and the order in which we choose the cookies doesn't matter. In a scenario like this, we use the choose function where 6 is the number of items we have to choose from and 2 is the number of choices we need to make.
For more explanations on using the choose function in questions like this, check out our video on combinations and permutations:
th-cam.com/video/M5GiD2pCQvM/w-d-xo.htmlsi=khn1jE5cGbfyZB5r
@@GMATNinjaTutoring I'll check that video out next! Have a hard time understanding why we need to choose 2 out of 6 since all 6 meet the criteria...
@@Tang586 this comes from the text of the question. We want to find out the number of ways we can choose "two [cookies] of each type." That means we need to find the number of ways of choosing two chocolate chip cookies and two oatmeal raisin cookies. We have six chocolate chip cookies available to choose from and as far as the question is concerned, all six cookies are identical, but we only want to choose two of them as that's the scenario presented in the question.
I hope that helps!
In the first question don't you have to get left+right shoes to become a matching pair?
I did (1/2)*(1/3)*(3/11)*(2/3)
1/2 for the right or left and 1/3 to be pink
3/11 to be pink and 2/3 to be left or right.
Total would be 1/36... Is my interpretation completely wrong?
Unless I've missed something in your calculation (1/2)*(1/3)*(3/11)*(2/3) = 1/33 which gets you the correct answer. However, I think you got a bit lucky in this question and I don't think it needs to be as complicated as this.
For the first choice, it doesn't matter whether we pick a left shoe or a right shoe. We need the shoe we choose in the second choice to be the other shoe in the pair, but we don't need to worry about that until we get to the second choice. For now, all we need to worry about is choosing a pink shoe. There are four pink shoes in our pile of twelve available shoes, so the probability of picking a pink shoe is 4/12 = 1/3.
For the second choice, we need this shoe to be the other shoe in the pair to make a matching pair. This means there is only one shoe from the remaining eleven that satisfies the scenario in the question. The probability of choosing this shoe is 1/11.
Combining these probabilities we get (1/3)*(1/11) = 1/33.
I hope that helps!
@@GMATNinjaTutoring you are right, I really did the math wrong hahahaha.
And yes, now I got it. Actually, I thought they were all the same shoes with different colors and it would not matter to pick one of the two remaining pairs for the first pink one he got, but that does not make sense. Thank you for clarifying.
Great tutorial, thanks
Thank you so much for watching!
In Q7 (Dogs Problem)
In statement 1, why did we multiply (L/9 × N/8) with 2?
We only have to find no. of ways to choose 1 Lab and 1 Non-Lab, without caring about the order. That's equivalent to LC1 × NC1 (combination)
It's when we multiply it with 2 or 2! that we start incorporating order (permutation)
Am I thinking of this in the wrong way?
I think you might be mixing up probabilities and combinations. If we were trying to find the number of ways you could choose one labrador and one non-labrador, then your method would be totally correct. However, Charles was trying to find the *probability* of choosing one labrador and one non-labrador. One of the ways of doing this is to find the probability of all the different ways of choosing one labrador and one non-labrador, and it turns out there are two:
(1) Labrador first, followed by a non-labrador
(2) Non-labrador first, followed by a labrador
To find the overall probability of choosing one labrador and one non-labrador in any order, we need to separately find the probability of options (1) and (2) and add them together. When Charles listed (L/9)*(N/8), that gives the probability of option (1). The probability of (2) is (N/9)*(L/8), which happens to give exactly the same answer as option (1). This means that instead of adding the probabilities of options (1) and (2), Charles was able to find the total probability we wanted by multiplying the probability of option (1) by 2.
I hope that helps!
@@GMATNinjaTutoring Thanks a lot for helping me understand the concept better. You guys are awesome!
For question 8, how do we find 1/8? In other words, how can we say that out of 80 numbers, 10 of them will be divisible by 8?
The key here is that the 80 numbers are consecutive integers.
To help understand the intuition, imagine that you have 10 consecutive integers. How many of those integers are even? If you're OK with that last question, try this one: how many of those 10 consecutive integers are multiples of 5? Play around a bit with the numbers until you see how the concept works.
Now, apply it to Q8: if you have 80 consecutive integers, how many of them are multiples of 8? Every 8th number must be a multiple of 8, and in this case, we have a quantity of consecutive integers (80) that's divisible by 8. So exactly 10 of them must be mutliples of 8.
The same concept also appears in this video on counting and sequence questions: th-cam.com/video/6GhWvbSTraw/w-d-xo.html&pp=gAQBiAQB. It's not necessarily the heart of the video, but you'll see echoes of this same idea there.
I hope that helps!
in Q6 is it correct to say: one of them can choose either the pie or not so 4/4 and the others must order the pie so 1x(3/4)^3 ?
In Chocolate Cookies question, how we can assume that all one type of cookies are not identical ? Although by solving them identical , answer is not in options . Question can be reworded properly to avoid confusion
This question asks us for the probability that we select two cookies *of each type* . It is the *type* of cookie that matters, not any other property of the cookies. This means that we don't need to assume the cookies are or are not identical. The answer would be the same in either case.
As long as all the chocolate chip cookies share the property of being chocolate chip cookies, it doesn't matter whether one chocolate chip cookie is identical to another in any other respect (and we could say the same thing about the oatmeal raisin cookies). As long as we can identify the type of cookie (chocolate chip or oatmeal raisin), we can solve this question using the method demonstrated in the video.
I hope that helps!
A query regarding question. 9 : Is the following method that I used correct? -
Probability of selecting 2 chocolate chip cooking consecutively and then 2 oatmeal raisin cookies consecutively : (6/10)*(5/9)*(4/8)*(3/7) = 1/14
Now, we can arrange the selection of 2 chocolate cookies and 2 oatmeal cookies in = 4!/(2!*2!) = 6 ways
Now we multiply 6 * (1/14) = 3/7
Would be thankful if you can validate this.
I thought of the exact same way , but i dont understand the part after we get the probability to be 1/14 .
Mind explaining please ?
@@abdur5908 Hi, after getting 1/14, we have to arrange the selection of 2 chocolate cookies and 2 oatmeal cookies. Think of it this way-
C- chocolate chip cookies
O- oatmeal raisin cookies
We have to select 2 of each type -
C,C,O,O
However, we also have to consider other cases such as O,O,C,C and C,O,C,O.
We can do this by multiplying 1/14 by 4! (Ways to arrange 4 items) and then by dividing 4! by 2!×2! (We do this because we have to prevent double counting cases, since chocolate cookie 1 does not differ from chocolate cookie 2)
@@pranavagarwal5605 OMG thank you so much! I was wondering why my approach was not getting me 3/7, totally forgot to account for selection order.
In question 8, would the 1/8 chance of having a number be divisible by 8 not be encompassed in the probability of drawing an even number? ex. 40, divisible by 8 and a even number, would that not be encompassed?
In this question, we're asked for the probability that n*(n-1)*(n-2) is divisible by 8, if we choose n randomly from the numbers 21-100.
In the first part of the solution, Charles asked how many ways could we choose an even number for the smallest value in n*(n-1)*(n-2). In other words, Charles asked how many ways could we choose an even number for the value of n.
In the second part of the solution, Charles asked how many ways could he find a value of the middle number in n*(n-1)*(n-2) that was a multiple of 8. In other words, Charles asked how many ways could we choose a multiple of 8 for the value of (n-1), which would give us an odd number for the value of n.
Because he was asking about different values for n in each case, the probabilities do not overlap.
I hope that helps!
For the last question, why can't we take (6/10)x(5/9)x(4/8)x(3/9) approach? But then that will have to be multiplied by by 4! I believe to account for the sequence, but that messes up the answer to be greater than 1. Can somebody help with this approach?
The issue with your approach is that you treat the two chocolate chip cookies and the two oatmeal raisin cookies as different cookies. In your process, if you pick chocolate cookie #1 first then chocolate cookie #2 that creates a different outcome than if you pick chocolate cookie #2 first then chocolate cookie #1. In this question, the chocolate cookies are interchangeable, so the two outcomes listed above would not be counted as two separate outcomes.
To fix this, you can take your final answer and divide it by 2!*2! to resolve the problem. This gives the correct final answer of 3/7.
I hope that helps!
These are amazing videos, thank you so much!
For question 7, part 2, since you multiplied part 1 by 2 to account for any order, isn't the same required for part 2? so 2(n/9*(n-1)/8)?
You're welcome! I'm so pleased you're enjoying the videos.
To answer your question: we should *not* multiply the expression we found when looking at statement (2) by 2.
When we looked at statement (1), we found (L/9)*(N/8) by considering what happens when we choose a labrador first and a non-labrador second. The reason we needed to multiply this by two is that this expression doesn't describe all the ways of choosing one labrador and one non-labrador. We also needed to consider what happens when we pick a non-labrador first and a labrador second, which gives (N/9)*(L/8). While this expression is slightly different from the first, it simplifies to the same thing so Charles could multiply (L/9)*(N/8) by 2 to give the final expression 2*(L/9)*(N/8) > 1/2.
When we looked at statement (2), we considered what happened when we picked a non-labrador and then picked a second non-labrador. We don't have a second situation to consider as there's no other way of choosing two non-labradors. Once we find (N/9)((N-1)/8) > 1/10, we have the situation described fully and there's nothing to add.
I hope that helps!
in case 2 at 21:50 timing 20 chef at conference can be male or women how it is sufficent?
If there are only 20 chefs at the conference, then the probability of choosing a chef is 20/75 = 4/15. This number is less than 1/3, so the probability of choosing a male chef is at most 4/15 (if there were no female chefs at all).
Since the probability of selecting a male chef must be at most 4/15, we know it is less than 1/3. This is why statement (2) is sufficient to answer this question.
I hope that helps!
I think in Q6 we have calculated the no. of pies for 4 customers and not exactly 3 of the customers, as we multiplied x4 and it should be done x3? Could someone pls help?
Think of it this way: imagine that the 4 customers walk into the restaurant one at a time. Let's call them customers A, B, C, and D.
Four outcomes are possible:
1. Customer A doesn't purchase a pie, but the other three do. Probability = 1/4 * 3/4 * 3/4 * 3/4 = 27/256.
2. Customer B doesn't purchase a pie, but the other three do. Probability = 1/4 * 3/4 * 3/4 * 3/4 = 27/256.
3. Customer C doesn't purchase a pie, but the other three do. Probability = 1/4 * 3/4 * 3/4 * 3/4 = 27/256.
4. Customer D doesn't purchase a pie, but the other three do. Probability = 1/4 * 3/4 * 3/4 * 3/4 = 27/256.
We could add these four probabilities together, and that's totally fine. But by now, you've probably noticed that the probabilities are identical for all four outcomes. So it's a convenient shortcut to just multiply by 4 instead of adding. Either way, we get the same answer (27/64), but we get there a tiny bit faster by multiplying.
I hope that helps!
@@GMATNinjaTutoring yes thank you so much ❤️
Q1 is misleading with the matching pink pairs. If you have to match left and right shoes, like you intuitively do with shoes, you get a second prob of 2/11 not 3/11. Logically 2/33 makes more sense as final solution.
The question tells us that Harry has to choose two shoes without replacement, so the first show will not be put back into the pile with the other shoes. Harry has a 4/12 = 1/3 probability of the first shoe he chooses being pink and, as you say, Harry has to match the second shoe to the first shoe in order to make a 'matching pink pair of shoes.' This means there is only one shoe in the pile of 11 remaining shoes that would match the first shoe Harry chose and give him a matching pair, so the probability for the second shoe is 1/11. We now have an overall probability of 1/3 * 1/11 = 1/33, which means (D) is the answer to this question.
I hope that helps!
For question 5, I did not understand why did you multiply to get the total outcomes (denominator = 4*3*3=36) instead of adding them?
Hm, this is one of those things that's hard to explain in writing, but I'll do my best.
On Q5, each "outcome" consists of one number from set x, one from set y, and one from set z. The idea is that FOR EACH potential selection from set x, you could have 3 different options from set y, and 3 more from set z. If that language ("for each") helps you grasp that multiplication would work, awesome.
If not, let's imagine that you choose a 2 from set x. Here are all of the possible outcomes with a 2 from set x:
2 5 7
2 5 8
2 5 9
2 6 7
2 6 8
2 6 9
2 7 7
2 7 8
2 7 9
We can repeat this process for every possible value of x (3, 4, or 5), and it would all look very, very similar.
Does that help you see why we would multiply? Sorry if that isn't clear -- this is the sort of thing that lends itself better to a picture than to a bunch of writing in the TH-cam comments section. :)
I hope that helps a bit, and have fun studying!
For Q8, the question asked for the probability of (n)(n-1)(n-2) being divisible by 8 if n is in the range of 21-100 inclusive. Slotting in the numbers, i found that in every 10 numbers, 6 are didvisible by 8. so in 80 numbers, 48 are divisible by 8 therefore isn't the probability 48/80 giving 3/5? i'm finding it hard to understand
It sounds like you just tried a sample of 10 numbers (something like 21-30) to see what happens? Try a sample of 16 or 24 numbers, and see if that changes your analysis at all. :)
The solution shown in the video is probably the quickest way to get to the answer, but I wouldn't worry too much about this one if you struggle to understand that particular process. This is a TOUGH question, with some funky little quirks to it. It's not the sort of thing that's going to hold you back from a really great score, since the concepts aren't frequently tested quite like this, and the question is really hard.
So play with it a bit if you'd like, but don't lose any sleep over it.
I hope that helps a bit!
@@GMATNinjaTutoring Giving that i like to leave no stone unturned in regards to studying, i actually wrote out the entire set and found 50 numbers divisible by 8 which is naturally 50/80 hence 5/8. i'll try not to pass out from embarrassment now :)
@@frederickkingjnr Oh wow, that's dedication! No need to be embarrassed -- it's a hard question, and I think it's cool that you're willing to spend that much time exploring it. I'm usually too lazy to do that sort of thing. ;)
Have fun studying, Frederick!
it is fun learning with you ,you are very cute
Aw, I'm blushing. Thank you!
In Q3, if we expand the solution then P(1) + P(2) + P(3) + P(4) where P(1) atleast one of them buys beer and son on.
P(1) = 3/4 *1/4 *1/4 *1/4=3/256
P(2)=9/256
P(3)=27/256
P(4)=81/256
Adding this should also give 255/256 but it comes out to be 120/256. Can you pls see where am I going wrong?
P(1) would have to be multiplied by 4 (4C1), P(2) by 6 (4C2) and P(3) by 4 (4C3) to get all combinations of customers of beers and not just 1/2/3 customers of our choice. You could take the 6th problem as a reference is this is difficult to grasp.
Hi @GMATNinja, thanks for the video. I had a doubt in Q8. This is the logic I put. For n(n-1)(n-2) to be divisible by 8, either n or n-1 or n-2 has to be divisible by 8 and since our range is 21 to 100 inclusive, I took the following 10 values of n that makes it divisible by 8; 24,32,40,48,56,64,72,80, 88, 96; similarly for (n-1) to be divisible by 8, I took the following values of n: 25, 33, 41, 49, 57, 65, 73, 81, 89, 97 and similarly for (n-2); I took: 26,34, 42, 50, 58, 66, 74, 82, 90, 98. As per this logic, we have 10 of each i.e. 30. So 30/80 = 3/8. Can you explain where I am going wrong with this approach. I am still confused by the explanation in the video.
Also, for Q9, is there a trick or a hint in these kind of questions that we have to Combination instead of going for the regular way of counting probability, before you started to explain, I did (6/10 x 5/9 x 4/8 x 3/7)x4 and landed on 2/7. In the exam, is there a hint I can look for to know that its a combination style question.
Thanks.
For the PRODUCT of n(n - 1)(n - 2) to be a multiple of 8, we don't necessarily need any of the individual terms to be a multiple of 8. For example, if n is merely a product of 4 (but not 8), (n - 2) would be even, and the product would be a multiple of 8 -- even if neither n nor (n - 2) is a product of 8. This would be the case if n = 20 or n = 26.
That's why the solution shown in the video works better. The product of the three terms will be a multiple of 8 as long as n is even -- regardless of whether n itself is a multiple of 8. If n is odd, the product will be a multiple of 8 if (n - 1) is a multiple of 8.
I hope that helps!
Is the second question a 'without replacement' question? The question asks us to find out the probability that the customer sits on a diff chair on each visit. Shouldn't the no of chairs available in that case decrease in the similar manner as in question 1?
Please can you let me know if I am reading too much into the question?
In Q2, picture what happens whenever Bransen walks into the barbershop. All of the chairs are still there, right? Yes, we're looking for the probability that Bransen sits in a different chair on each visit -- but he could still get assigned to ANY of the three chairs on each visit.
So this is a "with replacement" question, if you like that terminology. The chairs aren't disappearing after Bransen sits in one of them -- they're staying in the barbershop, no matter what.
I hope that helps!
Very basic question but how do you different between mutually exclusive and independent in probability - when to add / when to multiply?
Good questions. I deliberately avoided discussing some of those "textbook" terms (independent vs. dependent, mutual exclusivity) in detail in the video, just because the terms themselves almost never appear in actual GMAT questions, and I don't think that a textbook understanding of them helps much.
"Mutually exclusive", for example, shouldn't inspire much fear if you do see it. It just means that two things can't happen at the same time. So if you flip a single coin, you can't get both heads and tails simultaneously. Again, you're unlikely to see the term itself on the GMAT, and if you do, just apply common sense to the context of the question, and you'll be fine.
"Independent" just means that the outcome of one random event won't affect the outcome of another. For example, if you flip a coin and roll a 6-sided die, the result of the coin flip won't affect what happens on the die roll -- and since those two things are independent, you can multiply to get the probabilities to get the likelihood of any paired outcome (e.g., the probability that you get a tail on the coin flip and an even number on the die roll).
Don't get me wrong: in real-life statistical applications, independence is an absolutely crucial concept that can get remarkably thorny. But you're very unlikely to see the word "independent" on a GMAT probability question. And if you do, just understand that it means that two (or more) events don't affect each other, and you can multiply to get the probability that both things happen. That's really all you need to know for the GMAT or EA, practically speaking.
I hope that helps!
For Q5 - The numbers 7 and 5 are repeated in two sets so should we not consider different combinations? If yes, that would increase the number of desired outcomes
The repeated numbers are taken into account in the solution in the video. Since the sets have 4, 3, and 3 items in them, respectively, there are 36 total outcomes, regardless of the repeated numbers.
The repeated numbers don't change the number of "desired" outcomes at all, either. Starting at the 25:30 mark, I write down the ONLY four ways to get a sum greater than 19, and each of those outcomes is written in the same order: one number from set X, one from set Y, and one from set Z. The fact that 5s and 7s appear in more than one set doesn't affect the overall count of desired outcomes, since we're selecting exactly one number from each set.
I hope that helps a bit!
Hi, thanks a lot for the videos. Been watching your videos since last week, and it's amazing.
For Q7, I am still not quite sure why you multiplied (L/9)(N/8) by 2? Didn't understand your explanation tbh. Isn't the statement specifically stating that the first pick is a lab and the next one is a non-lab?
The first statement tells us "The probability of picking one labrador and one non-labrador is greater than 1/2." If we pick a non-labrador with the first selection and a labrador with the second selection, we have still picked "one labrador and one non-labrador." This means the statement tells us that the order of selecting the labrador and non-labrador does not matter.
Since the order does not matter, we have to consider the probability of selecting a labrador first and a non-labrador second and the probability of selecting a non-labrador first and a labrador second. To do this we can take the probability of selecting a labrador first and a non-labrador second (L/9)(N/8), and we can multiply this by 2 to take into account the probability of selecting a non-labrador first and a labrador second.
We can do this because the two probabilities end up being equal, even though the order of the selection is different. For a much more detailed explanation of why this is the case, check out the explanation to Q6.
I hope that helps!
For selection at random we use the combination formula and for selection without replacement we use the probability method is it so?
I'm not quite sure what you mean. The majority of probability, combinations, and permutations questions deal with random selection in some way or another, so we definitely couldn't say that we always use the combination formula when something is selected at random. And I'm not sure what you mean by the probability "method" -- it's an entire branch of mathematics, and you'll use different methods to solve different probability questions, as discussed in the video.
Maybe our video on combinations and permutations might help a bit? th-cam.com/video/M5GiD2pCQvM/w-d-xo.html. At the very least, it will hopefully help you understand when to use the combination formula, and when to avoid it.
I hope that helps a bit!
On question 8, can’t we think of this as the below:
n has a 50% chance of being even, in which case (n)(n-1)(n-2) will definitely be divisible by 8. (1/2) x (1/1) = 1/2
n also has a 50% chance of being odd, in which case there is a 1/8 chance that (n)(n-1)(n-2) is divisible by 8.
(1/2)x(1/8) = 1/16
1/2 + 1/16 = 9/16… shouldn’t this be the answer?
In other words, why don’t you have to multiply the 1/8 by 1/2, since this scenario requires n to be odd?
As a follow up, it seems like your solution would be double counting the “desired outcomes” (instances where (n)(n-1)(n-2) is divisible by 8) when both n is even and (n-2) is divisible by 8. These desired outcomes are already accounted for, since any time n is even we get a desired outcome
@@obm13 Great question! I don't know that I'll do a very good job of answering it without a whiteboard or a pencil, but I'll do my best.
You're spot-on with the first part: if n is even, (n)(n - 1)(n - 2) will definitely be divisible by 8. So that gives you 1/2, exactly as you mentioned.
For the next part, I think it helps to focus on (n - 1). There's a 1/8 probability that (n - 1) will be divisible by 8. Here's the thing: if (n - 1) is a multiple of 8, then n and (n - 2) will DEFINITELY be odd. That's why you don't need to multiply 1/8 by 1/2 in this situation -- they aren't independent events, since n will always be odd if (n - 1) is divisible by 8.
Does that help at all?
Got it. So because there is no scenario in which (n-1) is divisible by 8 AND n and (n-2) are even, we don’t need to worry about double counting with the top part. Makes sense - thanks
@@obm13 Oh good, I'm glad that my response made sense! Probability is often one of those things that doesn't translate well in TH-cam comments. :)
Have fun studying!
It is easy to see in this question that the probability is at least 1/2. Once you realise that multiples of 8 within the given range will also satisfy the condition, it becomes obvious that probability is greater than 1/2, but not as high as 7/8. So, 5/8 can be chosen with some confidence.
Had the options been closer, this question would have taken more time.
GMAT ninja team, thank you for all the efforts that you have put in making this series. Surely it is helpful in understanding the topics, but I think a more important function that your creation serves is to boost confidence and faith in test-takers! You have been an inseparable part of my GMAT journey. Thank you again!
For Q7, could you not say since the p(non lab + lab) is > 1/2, any other combination is going to have a probability of < 1/2 , including getting 2 labs
Yes, exactly. Another viewer pointed out the same thing, and I was pretty embarrassed that I missed the loophole in my own question. The explanation in the video is definitely more complicated than it needs to be.
So you deserve a cookie for finding the faster way. Have fun studying, Quinn!
Clarina explains great!
On Q6 if the order of the customer ordering the 3 pies does not matter why do we multiply the probability 4 times (raising to the power of 4).
and any examples where we do similar calculations but each outcome could be different, meaning that instead of raising to the power of 4 (4 outcomes the same), we would multiply different probabilities outcomes together, and why is it that we need to multiply them together? (I can only think of examples like if day one 3/4 order pies and second day 2/4 order pies, etc).
On Q6, I wouldn't recommend thinking in terms of whether "order matters" or "order doesn't matter." That's a crucial distinction for combinations and permutations, but I think it can throw you off in this situation.
You can think of Q6 this way: four customers will walk in the door, one after another. If you'd like, you can call them customers A, B, C, and D. Beginning at around the 30-minute mark of the video, we establish four scenarios in which exactly three customers order pie.
In the first scenario, for example, A, B, and C all get pie, and D gets none. What's the probability of this scenario occurring? Well, the probability that A gets pie is 3/4 -- and the same is true for B and C. There's a 1/4 probability that D does NOT order pie.
If these four are all independent events and we need all four to happen, then we just multiply the probabilities together: (3/4)(3/4)(3/4)(1/4) = 27/256. More broadly: whenever you're trying to compute the probability that two (or more) independent events BOTH happen, then you'll always multiply.
Q6 gets a bit more complicated because we have four different scenarios to cover, but hopefully this helps you understand why we multiply to get the probability for each scenario. And again: this isn't something that gets tested very often, so don't lose too much sleep over it.
@@GMATNinjaTutoring Thanks for the reply, I was (still am) new at probability and didn't understand much but after finishing the video I had a better grasp. I see that when calculating the probability of an outcome we also need to account for the different possible outcomes, like: D not getting pie, then C not getting the pie and so on.
@@GMATNinjaTutoring BTW watching the Q6 part again, when you do the (3/4)(3/4)(3/4)(1/4) I want to clarify it's (3/4)(3/4)(3/4)(1/4) x 4 and not (3/4)(3/4)(3/4)(1/4) to the power of 4, while explaining Charles clearly says "multiply this by 4" but the writing on the board the 4 is small and on the top right so also looks like a power of 4. I forgot to clarify that part I guess that's where my confusion of multiplying the final result 4 times by itself (to the 4th power) is coming from.
@@GMATNinjaTutoring I still dont get why didnt we stop at (3/4)(3/4)(3/4)(1/4) = 27/256. why did we have to multiply by 4? why we didn't do this for question 3?
@@BSA77 For Q6, let's go back to the idea that we have four distinct customers who come in. Let's call them customers A, B, C, and D again.
In this question, we want the probability that EXACTLY three customers order pie and EXACTLY one customer gets none. (Mmm... pie.) There are four ways for this to happen:
- A, B, & C get pie; D gets none. Probability of this outcome = (3/4)(3/4)(3/4)(1/4) = 27/256
- A, B, & D get pie; C gets none. Probability of this outcome = (3/4)(3/4)(1/4)(3/4) = 27/256
- A, C, & D get pie; B gets none. Probability of this outcome = (3/4)(1/4)(3/4)(3/4) = 27/256
- B, C, & D get pie; A gets none. Probability of this outcome = (1/4)(3/4)(3/4)(3/4) = 27/256
So what's the total probability that EXACTLY three customers order pie? We could just add these four numbers together. Or because all four probabilities are the same, we could just multiply by 4 instead. That's how we get 108/256 = 27/64.
Q3 is asking for something completely different: the probability that AT LEAST one of the customers will order a beer. Sure, you could take a vaguely similar approach as we did to Q6, but it creates a big mess. You'd have to calculate the probability that EXACTLY one customer orders a beer -- which is the same process as Q6. But then you'd have to calculate the probability that EXACTLY two customers order beer. And then the probability that EXACTLY three customers order beer. And so on.
That's why we take a completely different approach to Q3. Whenever you see the phrase "at least one" in a probability question, it's much easier to think in terms of the complement. In Q3, we're asked for the probability that at least one customer orders a beer; I'd much prefer to calculate the probability that ZERO customers order a beer, as shown in the video.
But again: Q6 is pretty tough, so don't lose too much sleep over it.
I hope that helps a bit!
it's a fantastic video loved the way you presented it in such a simple way. Are there such questions available so that I get a little hands-on with it. Thank you so much
Thank you so much for the kind words, Mansi! I'm glad that the video helped a bit.
Sadly, we haven't released any practice sets to accompany the videos yet -- sorry for the bad news. The good news is that all of our lessons are rooted in the types of questions you'll see in the official guides and official practice tests. So hopefully, you'll see echoes of this video as you work through the official materials.
Have fun studying!
@@GMATNinjaTutoring thank you so much for your response. This is a great help for me I will definitely look into it.
Hi Charles - Question for QN#3. When I add the probability, I also end up with 120/256 which is different from using the complimentary formula, which gives 255/256, why the discrepancy? Thanks!
Hm, I can't figure out how you would get 120/256. For starters, it should strike you as an implausible answer. The question is asking for the probability that at least one of the four customers orders a beer. If the probability that each INDIVIDUAL customer orders a beer is 0.75, then the correct answer must be even larger than that, since the bartender has four opportunities to get somebody to order a beer.
The only solution I can think of that involves adding probabilities is pretty darned miserable. You would basically have to write out all of the different ways that at least one person orders a beer, calculate the probability for each of those scenarios (there are 15 of them!), and then add them. That's terrible, and won't be a good path to happiness.
The takeaway: when the question asks you for "at least one" of something, you're much better off thinking in terms of the complement. It makes your life MUCH easier.
I hope that helps a bit!
I gave up on the questions number8&9 too difficult for me :( I wish there is simpler formula to memorize this kind of questiion.
Don't worry too much about #8 & #9! As we say throughout the video, probability questions are a small proportion (2-3%) of the total on GMAT quant. Unless you're chasing a 51Q for some reason, you'll be just fine if you can handle the fundamental concepts that appeared in the first few questions in the video. You're very unlikely to see something as hard as #8 or #9 on the actual exam -- and if you do, getting those questions wrong definitely won't ruin your day.
I hope that helps a bit!
On Q8, about the p(multiple of 8) from an (odd)(even)(odd) scenario is 1/8 because in theory if there's only 1 even number in order to be divisible by 8 it can only be multiples of 8 (8, 16, 24, 32, etc) so 1 every 8 numbers. But wouldn't in this case be 1/10 or 8/80 because in theory there will be 8 numbers that can be multiple of 8 inside the 80 integers?
I'm not 100% sure that I understand your question correctly, but it looks like you might be trying to answer something different than what Q8 is asking. Q8 is asking for the probability that the PRODUCT of three consecutive numbers is even; it sounds like you might be trying to calculate the probability that ONE randomly selected number is a multiple of 8.
If you're trying to answer the latter question, the answer would actually be 1/8 (not 1/10), since every 8th number is a multiple of 8. For more on this type of thing, check out this video on "counting": th-cam.com/video/6GhWvbSTraw/w-d-xo.html.
You're fairly unlikely to see anything quite as tough as Q8 on your actual exam, but the solution given in the video is correct. It's a pretty nasty little question, so don't lose too much sleep over it. :)
I hope that helps a bit!
@@GMATNinjaTutoring Thanks for explaining where my mistake might come from, seems that I over fixated on the fact that it was only 80 integers. Thanks again and amazing series.
@@kelvincheng3103 Thank you so much for the kind words! And sorry for my dodgy handwriting. ;) Have fun studying!
I don't think I ever answered "E" (together not suff) when the correct answer was "D" (both alone are suff) before Q7 on this video haha
Got 4/9 correct, don’t think that’s very good but had a friend who ended up with 760 and said the strategy for him was to guess any probability questions haha and ended up at Wharton, so that’s something we can all consider … knowing the basics and then if know it’ll take more time, let’s just move on!
Exactly! And that's part of the point of the video. It's fun to mess around with the harder versions (I had fun with the video, anyway; your results may vary), but unless you have some really compelling reason to chase a 50+ on quant -- and nothing more productive to do with your time -- all you really need to do is focus on the very very basics, and don't worry about the rest.
Only around 2% of GMAT and EA quant questions involve probability. So you don't want to miss the easy ones, but it's fairly unlikely that you'll see any wacky, hard variants. And if you do, they won't hold you back from scoring well into the 700s. :)
Q8: Probability of multiples of 4 should be 1/4 & NOT 1/2.
Total probability therefore being: 1/4 + 1/8 = 3/8
I think you're thinking of the probability that a single integer is a multiple of 4. In Q8, if we were just looking for the probability that n itself is a multiple of 4, you'd be correct: that probability would indeed by 1/4.
But in Q8, we want the probability that the PRODUCT of three consecutive integers -- (n)(n - 1)(n - 2) -- is a multiple of 8. And here's the thing: if n is even, that product will always be a multiple of 8. Why? Well, if n is a multiple of 4, then (n - 2) is even... and the product of those two numbers will be a multiple of 8. But if n is a multiple of 2 (but not 4), then (n - 2) must be a multiple of 4... and the product of those two numbers will still be a multiple of 8.
So as long as n is even, the product of the three integers will be a multiple of 8. And that's why the probability is 1/2 + 1/8 = 5/8.
I hope that helps!
@@GMATNinjaTutoring Bingo!! You are a genius.... Thanks a ton... Respect your passion & enthusiasm
@@AmitSharma-zh5ju Haha, thank you so much! Have fun studying. :)