U wouldn't know how perfectly times this is, I have a test on permu's tmrw and wasn't bothered to do work so ended up procrastinating and finding Ur videos, keep up the great work dude. 👍
I graduated 1982 with EET degree. I try to help my daughter, she is taking Algebra II. At senior age remembering math is little difficult. I watch your video several time and pausing to understand it. Thanks.
Thank you, so much. ❤i wasn't able to understand some questions in statistics for economics. These basic solutions give me ideas on how to solve them. love it.
A fanstastic collection of probabilty questions which span several concepts. Q14. Ironically combination lock means order does not matter. They should really call them permutation locks.
The part with the cards is one of the hardest because i dont play that game since it's really for grownups. I keep messing up since i dont understand the fundamentals of the game. But i still understood some of the materials so thank you.
Thank you so much Mario! I just have one question: why do we subtract 1 before we calculate?(number 8, salade bar question), every question that contains 2 possibilities (yes or no), do we need to solve in the same way like you did? (2^n - 1)
Have you done any examples where you are looking for when 2 objects are next to each other. Ex: there are 12 balls in total, 6 are red. What is the probability of not having the 2 balls next tot each other
This question is a bit confusing but it said we're trying to figure out the order which means we remembered the digits but NOT the order. So ORDER for 4 digits that's 4!. We have that with one chance= 1/ 24. If we're trying to find out the digits, we need to choose from 10 out of 1.
well one i didn't find anywhere else is when there are repeating letters in a word .... the questions is......: 3 letters are to be chosen from the word BIOLOGY. Find the number of ways that it can be done if it contains both of the 'O's .
Another question for you Mario. Why can't we solve question 10 like we did with 9? So (7C4 * 3C1) / 10C5 ? Since there are 10 tries, 7 successes and 3 failures. But this gives 41.7% which is not our answer. But why though?
40:17 Can someone please explain why it's 4P4 and not 10P4 please? Isn't it possible to have 0-9 digits for the 5 digit lock? If I need to fill in 4 spaces, but I have 10 number options, why is it not 10P4? Thanks in advance!!🙏
I was thinking about this too. Cause there are 10 possible digits which are 0-9. Assuming that you know the middle digit, using fundamental counting principle, it should be 9•8•1•7•6 = 3024. And one chance of getting it correct would be 1/3024.
I think the answer for question number 7 is wrong. It should be 5*7*6*5. We have 8 choices( excluding 0 and 1) . The 8 digits are 2, 3, 4, 5, 6, 7, 8, 9. To have number less than 7000, the first digit must be either 2 or 3 or 4 or 5 or 6. So for the first digit we have 5 choices. For the second digit, we have 7 choices( from the initial eight choices, one digit has been taken for the first digit of the 4-digit number that we want to form). For the 3rd digit , we now have 6 digits to choose from( out of the 8 digits, we have used 2 digits for the first and second digits of the 4- digit number). For the forth digit, we are only left with 5 choices using FCP. We have a total of 5*7*6*5 chouces = 1050 choices.
Thanks for commenting…however I think you may have misread or misunderstood the question. It is a 4 digit number with no 0’s and no 1’s. The digits can be repeated. For a number less than 7000 the first digit has to be a 2,3,4,5, or 6 which is 5 choices. The second digit has to be 2-9 since we are excluding 0 and 1 which is 8 choices …same for the 3rd and 4th digits. So 5x8x8x8 should yield the number of possibilities. I hope that helps clarify for you and other viewers.
Actually the question was not adequately "worded". From the concept or introductions of permutation, we assumed no repetition, so if there should be repetition of digits of digits, that condition must be added. What I solved is correct based on "repetition of digits is Not allowed". What you solved is also correct if "repetition of digits is allowed". But your question didn't specify that repetition is allowed. Anytime permutation is introduced or explain at the beginning of a lesson, the assumption has always been "repetition is Not allowed".
U wouldn't know how perfectly times this is, I have a test on permu's tmrw and wasn't bothered to do work so ended up procrastinating and finding Ur videos, keep up the great work dude. 👍
Glad I could help!
I graduated 1982 with EET degree. I try to help my daughter, she is taking Algebra II. At senior age remembering math is little difficult. I watch your video several time and pausing to understand it. Thanks.
You’re welcome!
How sweet you are.😊.it show your affection towards your daughter ❤
Sir, you are better than all my math professors at college at clearly explaining math concepts and showing how the formulas work
Glad to hear my videos are helping you!
Thank you for explaining these concepts in a simple and straightforward manner.
You're very welcome!
Thank you so much 😊
Mario rocks!!! 👏🥳
Thanks a lot sir!
Bless you for this. Wish you only good things in life ❤
Thank you, so much. ❤i wasn't able to understand some questions in statistics for economics. These basic solutions give me ideas on how to solve them. love it.
Glad to hear that!
You just made college way more easier and now I'm ready for finals. Thank youuu
I'm so glad!
Thank you you made me happy. The first time after many days I understand this topic. You made it simple and straightforward
Glad to hear that!
Best video ❤
Im trying to self study probability theory and really struggling with perms and combs probability questions. This video helped so much, thank you!!!
Glad it was helpful!
A fanstastic collection of probabilty questions which span several concepts. Q14. Ironically combination lock means order does not matter. They should really call them permutation locks.
tysm u saved my life
Wish me luck on the ACT Mario! I appreciate the help
wow... great choice of problems for this video! I feel like you chose ones that are good to use for ACT prep. thanks
Thanks Jayden! Glad you liked my choice of problems! Good luck on the ACT!
The part with the cards is one of the hardest because i dont play that game since it's really for grownups. I keep messing up since i dont understand the fundamentals of the game. But i still understood some of the materials so thank you.
i had always problem in this subject(probability)but now after long time im now understanding it thanks man your amazing teacher!
You are most welcome
Your video is so helpful. It helps me lot. Thank you and keep it up. ❤❤❤😊
Glad to hear that!
Your teaching is amazing.
Glad you think so!
Really helped, by the way.
Keep it up.
Thanks, will do!
u r the goat
very helpful , i would recommend it to anymore.
Great to hear!
thank you so much
You’re welcome!
Love your videos. Could you do some more ACT math videos? There is one in September and October. Thanks.
Have you seen my ACT video course playlist on TH-cam? I’ll see if I can film another ACT review sometime soon…
Amazing teacher!
Thank you! 😃
If you are my university I will get 100/100 because you are a good teacher some teacher have knowledge but no teaching skills🙏🙏
Thanks for your positive comment!
Thank you for your videos, they have been a great help. Do you have any videos on birthday probability problems?
Not yet!
thank you very much
You are welcome!
thank you so muchhhh
You're welcome!
Fantastic video
Thanks! 😃
excellent video thanks for sharing
Glad you enjoyed it
Thank you so much Mario! I just have one question: why do we subtract 1 before we calculate?(number 8, salade bar question), every question that contains 2 possibilities (yes or no), do we need to solve in the same way like you did? (2^n - 1)
pls video in sets
Amazing tips Mario! In number 2, what would the answer be if ties were possible?
Im wondering the same thing
This guys awesome
Thanks!
Nice work
Thank you! Cheers!
For a question like number 7 do we always assume that the numbers can repeat?
Hello, thank your for your video! For #3, why is it that we cannot use the fundamental counting principle and thus calculate it as 15 x 14 = 210?
You could do that if it was done by Permutation , but in this question the order doesn't matter so we use companiation
Have you done any examples where you are looking for when 2 objects are next to each other. Ex: there are 12 balls in total, 6 are red. What is the probability of not having the 2 balls next tot each other
@2:31 that made my day 😂😂
why HT and TH are considered two different outcomes eventhough they are same
In problem 5, what would have been the answer for (at least 1 Ace) ?
Idk for sure. But i think you could do total possibilities minus no aces. So 52C5 - 48C5 = 886 656.
Let me now if this is correct
my math exam is tomorrow and im binge watching these videos. u might get me through university! hahahaha
You got this!
For question 14, why the answer is not 1/10^4 ? We have 0-9, 10 10 numbers, to choose for 4 digits.
This question is a bit confusing but it said we're trying to figure out the order which means we remembered the digits but NOT the order. So ORDER for 4 digits that's 4!. We have that with one chance= 1/ 24.
If we're trying to find out the digits, we need to choose from 10 out of 1.
well one i didn't find anywhere else is when there are repeating letters in a word .... the questions is......: 3 letters are to be chosen from the word BIOLOGY. Find the number of ways that it can be done if it contains both of the 'O's .
Another question for you Mario. Why can't we solve question 10 like we did with 9? So (7C4 * 3C1) / 10C5 ? Since there are 10 tries, 7 successes and 3 failures. But this gives 41.7% which is not our answer. But why though?
40:17 Can someone please explain why it's 4P4 and not 10P4 please? Isn't it possible to have 0-9 digits for the 5 digit lock? If I need to fill in 4 spaces, but I have 10 number options, why is it not 10P4? Thanks in advance!!🙏
You know what the digits are you just forgot the order that they're in.
I was thinking about this too. Cause there are 10 possible digits which are 0-9. Assuming that you know the middle digit, using fundamental counting principle, it should be 9•8•1•7•6 = 3024. And one chance of getting it correct would be 1/3024.
Is there shortcut for the calculation
Sadly, no
saitama!, is that you?
Ahhh important to know aces aren't face cards for question 5, e
I think the answer for question number 7 is wrong. It should be 5*7*6*5. We have 8 choices( excluding 0 and 1) . The 8 digits are 2, 3, 4, 5, 6, 7, 8, 9. To have number less than 7000, the first digit must be either 2 or 3 or 4 or 5 or 6. So for the first digit we have 5 choices.
For the second digit, we have 7 choices( from the initial eight choices, one digit has been taken for the first digit of the 4-digit number that we want to form).
For the 3rd digit , we now have 6 digits to choose from( out of the 8 digits, we have used 2 digits for the first and second digits of the 4- digit number).
For the forth digit, we are only left with 5 choices using FCP. We have a total of 5*7*6*5 chouces = 1050 choices.
Thanks for commenting…however I think you may have misread or misunderstood the question. It is a 4 digit number with no 0’s and no 1’s. The digits can be repeated. For a number less than 7000 the first digit has to be a 2,3,4,5, or 6 which is 5 choices. The second digit has to be 2-9 since we are excluding 0 and 1 which is 8 choices …same for the 3rd and 4th digits. So 5x8x8x8 should yield the number of possibilities. I hope that helps clarify for you and other viewers.
Actually the question was not adequately "worded". From the concept or introductions of permutation, we assumed no repetition, so if there should be repetition of digits of digits, that condition must be added.
What I solved is correct based on "repetition of digits is Not allowed". What you solved is also correct if "repetition of digits is allowed". But your question didn't specify that repetition is allowed. Anytime permutation is introduced or explain at the beginning of a lesson, the assumption has always been "repetition is Not allowed".
I also did it with no repetition allowed so i was a bit confused at first
I think you made a mistake on question 12 it’s a combination therefore we cannot use the first formula you used.
First
I started feeling stupid after no.7, lecturers didn't teach this course at all.
Why make maths complicated with all these notations ?
I never understand
I just stopped watching at question number 7
We need u in lebanon🥲🫶