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.
Thanks a lot man, I have a test tomorrow and I was really worried I would not be able to pass it as I could not understand this topic very well, but you saved me! I am forever grateful🙏🙏
Professors and Teachers actually teach based on the length of time assigned to them and materials they try to give students at a short period of time makes students feel they aren't good
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)
This is one of the few topics where I feel like I am just plugging numbers to get the right answer but don't really understand fully the concept. I understand it just enough to barely get by but still feel unsatisfied when I get the "right" answer.
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?
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!
Crazy
15:46 This has made so much of a difference 😮 Also 22:54. My class does not use this notation but it makes so much sense and is easier to understand.
Yeah. What are you studying
Glad to hear the video helped you!
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 ❤
Your teaching is amazing.
Glad you think so!
Thank you for explaining these concepts in a simple and straightforward manner.
You're very welcome!
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!
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!
Mario rocks!!! 👏🥳
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
Thank you, this will really help get into mathcounts.
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!
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!
Excellent explanation!👏👏👏
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!
Best video ❤
This was really helpful
Amazing teacher!
Thank you! 😃
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.
Bless you for this. Wish you only good things in life ❤
Thanks a lot man, I have a test tomorrow and I was really worried I would not be able to pass it as I could not understand this topic very well, but you saved me! I am forever grateful🙏🙏
🙏
Glad it helped!
every time i am gonna have a summative i will go through this video! really helpful!!
I’m glad my video helped you prepare for your summative!
Really helped, by the way.
Keep it up.
Thanks, will do!
thank you so muchhhh
You're welcome!
Your video is so helpful. It helps me lot. Thank you and keep it up. ❤❤❤😊
Glad to hear that!
thank you so much
You’re welcome!
Amazing tips Mario! In number 2, what would the answer be if ties were possible?
Im wondering the same thing
Thanks a lot sir!
Am new here mwalimu,I really really like to learn more from you in entire mathematics topics.
I hope my channel can help you learn more math!
Thankyou so much❤
You're welcome 😊
tysm u saved my life
Wish me luck on the ACT Mario! I appreciate the help
very helpful , i would recommend it to anymore.
Great to hear!
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!
Professors and Teachers actually teach based on the length of time assigned to them and materials they try to give students at a short period of time makes students feel they aren't good
Thank you for your videos, they have been a great help. Do you have any videos on birthday probability problems?
Not yet!
Fantastic video
Thanks! 😃
mario you are as genius as Eienstein for sure you have a big brain
I wouldn’t say that…but I’m glad you like my videos.
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.
excellent video thanks for sharing
Glad you enjoyed it
You just made college way more easier and now I'm ready for finals. Thank youuu
I'm so glad!
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)
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…
u r the goat
Nice work
Thank you! Cheers!
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
@2:31 that made my day 😂😂
This is one of the few topics where I feel like I am just plugging numbers to get the right answer but don't really understand fully the concept. I understand it just enough to barely get by but still feel unsatisfied when I get the "right" answer.
This guys awesome
Thanks!
For a question like number 7 do we always assume that the numbers can repeat?
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
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
my math exam is tomorrow and im binge watching these videos. u might get me through university! hahahaha
You got this!
why HT and TH are considered two different outcomes eventhough they are same
Beta tuuu koi galat sums samajh leee😂 combination kaaa matlab hii tujhee malum nahi ayaa
pls video in sets
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?
Is there shortcut for the calculation
Sadly, no
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
saitama!, is that you?
Can anyone explain number 4 to me?
Ahhh important to know aces aren't face cards for question 5, e
I think you made a mistake on question 12 it’s a combination therefore we cannot use the first formula you used.
I started feeling stupid after no.7, lecturers didn't teach this course at all.
First
Why make maths complicated with all these notations ?
verified
I never understand
I just stopped watching at question number 7
We need u in lebanon🥲🫶
Thank you so much 😊
thank you very much
You are welcome!