for Q3. If you think about it, you may notice that for any number rolled on the first die there'll be exactly 1 number to complement it to 7. Which means that you can completely ignore the first die and say that it's the chance of rolling a specific number on the 2nd die which is solved in Q1. In Q4 you can simply add the missing 9 cards to the 13 hearts. it's pretty much demonstrated by the venn diagram in the video Q6. Similarly to Q3, we know that rolling a pair is 1/6. Which means that NOT rolling a pair is 5/6. All we have to do now is check what are the chances of not rolling a pair 3 times in a row. 3/6 times 3/6 times 3/6 = 0.5787 to NOT roll a pair in under 4 tries which gives us 1-0.5787=0.4213. My point is, don't focus on formulas. Think first and try to reduce a problem to something you already know. Analytical thinking > memory. That's kinda the way a mathematician is supposed to think anyway Most of these small probability and statistics questions don't require any formal training P.S. also if you happen to be familiar with simple combinatorics, most of the simple probability problems can be reduced to combinatorics problem: may not necessarily be as elegant but if you're familiar with that field better, it'll definitely be more efficient for you
I got them all. I tripped up on the Bayes Theorem problem at first, but then I got it. Number 6 can be done another way. "Taking less than four tries" is equivalent to "not taking four or more tries". Taking four or more tries can be calculated easily because that just means that the first three attempts are failures, so that would be (5/6)^3. Then not taking four or more tries would be the compliment of that, or 1 - (5/6)^3. That's equivalent to the answer in the video.
Another person mentioned it, but you should have both probabilites for false positive and negative for problem 10. They are not the same for almost every single real life application, and considering the difference between type 1 & 2 errors is crucial in statistics, which is the main application of probability theory!
for Q6 we can use the complement rule - meaning 1 Minus failure (which is 3 times of NOT getting a pair) - in this question (not all the time) this approach is easier than the geometric sequence one
9:30 We can also solve the question in this way. P(getting a pair(doubles) in < 4 tries) = 1 - P(not getting a pair(doubles) in < 4 tries) = 1 - Πi=1 to 3 P(not getting a pair(doubles) on try i) = 1 - (30/36)³ ≈ 0.4212.
For question 11, I just knew that one of the two questions would be in 1 of the 11 slots while the other is in one of the 10 remaining. Regardless of what slots the others are in. That means there are 120 possibilities of slot combinations for those two questions. I then calculated the number of possibilities that the questions are beside each other by considering number of permutations conditioning on the easiest question being in slot 1, then in slot 2, etc. and came up with 20. So it was a simple 20/110 or 2/11
Another solution method for 11 is that the easiest problem is either in position 1 or 11, in which case there is only one of 10 adjacent places, or in a position in between, in which case there are 2 of 10 adjacent places. So the probability is 2/11*1/10 + 9/11*2/10 which is again 2/11.
for question 3, you just need to think that for any result the first dice gets (lets call it A), theres only 1 number in wich A+B=6, so you need to find a single number at the second dice, 1/6
Here is a question that I once solved which is very good. Ques - Morse code uses dots and dashes, which are known to occur in the proportion of 3:4. Suppose there is interference on the transmission line, and with probability 1/8 a dot is mistakenly received as a dash, and vice versa. If we receive a dot, what is the probability that a dot was sent?
These are 2 more interesting problems (but much harder than those mentioned in the video) for those who managed to solve all the problems provided and want a challenge! Great video btw! Question 12: In the country of probaland, couples will have children until they have their first girl, at which point they will never have children again. On average, how many children does each couple have? Question 13: In the country of probaland, couples will have children until they have their first girl, at which point they will never have children again. If you were to choose a random person in this land, what is the probability they are a boy?
@@victory646813 is 0.5. Consider a fair game of coins (heads you win 1$, tails you lose 1$). There's clearly no strategy that gives a profit in the long run, thus no childbirth strategy can affect the overall ratio of boys to girls.
not difficult, but for Q10, the problem does not mention both true positive and true false have same conditional probability, we have to assume they are equal.
@@fenrir6723 anyway, it is still an assumption, we should state that as it is a problem. For it is a practical case, feel free to assume what you like.
@fenrir6723 That is actually a very dangerous assumption and is most certainly not true for almost every real-life case. For pathegon detection tests(like COVID and the given problem), false positives are somewhat acceptable while false negatives can be devastating, so they are purposely built to have an extremely low false negative rate but not that low false positive rates. For fingerprint recognition, it's more dangerous to have a false positive so it's built the other way around.
@@fenrir6723I agree that the problem should've stated both probabilities, as that is needed info for problem solving and in my opinion is not a trait that can be just assumed to be true.
I thoroughly enjoyed this video, and thanks for creating it! As for how I did, I was able to do #1 - #8, and #11. I did not try #10 because I don't like applying a theorem, in this case Bayes' Theorem, to a probability problem because it is usually a straightforward application. The real story is about problem #9. Although I correctly approached it by trying to take the complement of the probability that no two students shared a birthday, as you did, I treated the numerator of this probability to be a combination problem, i.e., in how many different ways I can pick 23 days out of 365. Also, for the denominator, i.e., the total #possibilities, I simply took 365. As I took a snapshot peek at your solution to see how I faired, I realized that the numerator could be arrived at by treating it as a permutation problem, and I was way off with the denominator. Without playing your solution, I tried to arrive at your expression for the probability that no two students shared a birthday, and interestingly, arrived at it using an approach that was quite different from yours. So, for the numerator, i.e., #favorable events, your expression made me realize that this is a P (365, 23) permutation. For the denominator, my reasoning was as follows. The 1st student can have 365 possibilities for birthdays and for each of these possibilities, the 2nd student can have 365 possible birthdays, and for each of the 365 x 365 possibilities across the first two students the 3rd student can have 365 birthdays, and so on, to give 365^23 possibilities across 23 students. So, the probability that no two students shared a birthday comes to P (365, 23)/365^23, i.e., the same as your expression. However, what blew me away is that your approach was completely different, and a remarkable one, I must say. I also realized that your approach is akin to sampling with replacement. To understand this better, I personally reframed the problem as: What is the probability of picking the same student at least twice when randomly and sequentially picking 23 students from 365 students with replacement? So, the real kicker for me is that I took a peek at your expression and arrived at it in a way that is very different from yours! I thoroughly enjoyed this detour for problem #9 :)
They are standard (and kinda easy) questions in probability theory so i got all of them correctly. If i havent been taught probability theory tho i wouldnt score more than 3-4 of them right
In question number 11 28:55, the numerator should be 10! * 2 * 10, the extra 10 multiplication because the boxed question group could shift itself to 10 different places... So there are 10 ways to arrange the group Q.1 and Q.11 across the blanks, 2! internal arrangements, and 10! Other element arrangements
I like your thinking but it would be 9! other element arrangements not 10!. Therefore what you are saying is that you would have 10*2*9! which is equal to 10! * 2 (which is how it is in the video) Hope that helps!
I got tripped up for question 4 because I’m more familiar with logic, so when you said or I thought it meant it could the one of the two or both, but it turns out that the or you mentioned was what we call XOR (exclusive or), meaning that it can’t be both. In retrospect it was pretty obvious that you were talking about that 😅! Great video.
It's not an exclusive or. It means hearts and/or a face card. He subtracts cards with both so as to not double count them as both face and heart cards.
The qns were fairly ez if you've done highschool math/A level math. The only one i struggled with was the binomial one cause i forgot how they work, also the last one was definitely not the hardest.
A different way of looking at that last one. There are 11 possible spots for the easiest question. In two of those (first and last) there is one of the remaining 10 spots where the hardest question can go and be next to it. In the nine other spots for the easiest question, the hard question can go in two of the ten remaining spots and be next to it. So 2/11 x 1/10 + 9/11 x 2/10 = .1818…
Alternatively, whenever the easiest and hardest question are next to each other they can be treated as a single object. Among 9 other questions there are 10! possible positionings, times 2 to account for the order of questions 1 and 11. Divide this by the total number of orderings (11!) for an answer of 2/11
14:31 For adults and children cant we do: (6/10) * (5/9) * (4/8) * (3/7) = 0.0714 Order doesn't matter, even if we pick children first like: (4/10) * (3/9) * (6/8) * (5/7) Can someone explain whats wrong with this approach?
The only thing you forget to account for is the position of the children and adults. Your solution assumes that the first 2 positions are ALWAYS children and the second 2 are ALWAYS adults. But what if there's a child in the first and third position? That fits the criteria as well AND has the same probability as your case. There are 6 different degenerate such cases, ({1,2},{1,3},{1,4},{2,3},{2,4},{3,4}), hence you need to multiply your calculations by 6 to get the final answer.
You're solving this via permutation rule. Permutations assume order matters automatically, so you have to do every single ordering of the adults and kids. Or you can use combinations like he did and not worry about that.
The First question is obviously 1/6 - you need 1 of the 6 sides. You DO NOT NEED the formula and if you do, you are Just making the solution redundantly complex and if used in algorithm, you would slow it Down by redundant calculation. Use LOGIC instead of patterns. - Tomas, senior software engineer
My answer for question 7 is 42.9% I'm so good at math EqualCount = 0 from random import randint trials = 1_000_000 for trial in range(trials): adults = 6 children = 4 for picks in range(4): if randint(1, adults + children)
Are these questions good for highschool student's exam??? I know only one formula which was taught that is, --> probability of event happening - probability of event not happening = 1
Question 11: I would explain like this to younger student to avoid as much as possible factorial, but your explanation is great. The pair easy/hard depends only on the placement of the easy question, which must always precede the hard one. There are 10 possible positions for the easy question (10 choose 1). Since the order can be reversed, the number of satisfying cases doubles. The denominator is determined by the number of ways to arrange two questions within a set of 11 (number of arrangements of 2 elements within 11). The nine other questions do not interfere directly on the probability. Therefore, P = (2x10)/(11×10)=2/11
Mathematically 8 was harder than 9 which was harder than 10 which was harder than 11. however actually punching in the numbers to answer question 9 made me want to throw my calculator against the wall.
I just still dont feel like i get the concept behind question 6...Why is it that the odds of hitting the pair decreases as the tries increases, why isn't it increasing? Shouldnt the more you try, the higher your chances of success? please i really want to learn...
In that question if you want to get the pair in a specific turn you need to be unsuccessful the turns until that turn that's why the probability shrinks by time. I am not a native English speaker I hope you understand
P1 is the probability that you roll a pair, 1/6. P2 is the probability that you rolled a pair (1/6) AND didn't roll a pair in the first one (5/6). Remember that we stop trying after we have rolled 1 pair. Does this help you?
Sir please my humble request is to upload next vedio regarding Intermediate value theorem , Mean value theorem, monotonicity, maxima and minima ,topic of Calculus
Ok I'm at question 5 rn and WHY DID YOU NOT SIMPLIFY THAT FRACTION je devienne fou THE LAST QUESTION!!! WHY NOT SIMPLIFY THOSE FACTORIALS INTO 2/11?!?!? me angey it's 4am I should go to sleep
I am so confused about number 9 right now. I figured I would try it and approached it by trying to count how many different ways there are for two students to share a birthday, which is quite simple, that being 22+21+20...+3+2+1, or 253. That represents the pairs of birthdays between those students. Each pair has a 1 in 365 chance to be the same. Basically you're just rolling two 365 sided dice 253 times and hope for at least one pair, or that's what I thought. But when calculating the probability (1-[364/365]²⁵³) my result was 50.0477154%, which is incredibly close to the answer you gave, but not quite there. But I double checked my inputs in the calculator multiple times and I can't find the flaw in my logic, yet it's still just ever so slightly off. Did I make a mistake? Did you make a mistake? Is it just complete coincidence? What's even more confounding to me is that if I do it to the power of 258 I'd get 50.728% (or 50.73% rounded) which is the result in the video. If you change the exponent by 1 the result is quite far away from that in comparison so it's not like you'd be expected to just get the answer eventually if you add or subtract 1 to/from the exponent enough. But 258 as an exponent makes absolutely no sense, so is it even more coincidence?
Yo I feel like question 4 is the hardest bc the way you’ve worded it. It says heart OR face, not heart and/or face, meaning any cards that are in both categories are nullified. This means it would be 10(hearts that aren’t face) +9 (faces that aren’t hearts)/ 52(total cards). This should leave you with 19/52.
Ultimately language is vague, but generally in probability in English, we think of OR as being the inclusive OR, with exclusive OR (XOR) needing to be explicitly stated.
I think you got 6 wrong. Imagine that after you threw the cube 4 times, and check if you got a double, in any of them. Getting a double in the first, does not negate your chance to get a double on the second Your calculations showed the chance of getting a double once wheen throwing 4 times. Just to prove my point, you can calculate the opposite, which is (5/6)**4, and see that your answer plus it's opposite is less than 1
His calculations showed that P(1) + P(2) = 1/6 + 5/6 * 1/6. This is the probability that the first or second roll is a double. We multiply the second roll by 5/6 bc that's the probability the first roll wasn't a double. We need that bc we only need 1 double before we stop rolling. Also, you can calculate this using 1 - P(only rolling non-doubles 3 times), which is 1 - (5/6)**3. This will give the same outcome as in the video.
for Q3. If you think about it, you may notice that for any number rolled on the first die there'll be exactly 1 number to complement it to 7. Which means that you can completely ignore the first die and say that it's the chance of rolling a specific number on the 2nd die which is solved in Q1.
In Q4 you can simply add the missing 9 cards to the 13 hearts. it's pretty much demonstrated by the venn diagram in the video
Q6. Similarly to Q3, we know that rolling a pair is 1/6. Which means that NOT rolling a pair is 5/6. All we have to do now is check what are the chances of not rolling a pair 3 times in a row. 3/6 times 3/6 times 3/6 = 0.5787 to NOT roll a pair in under 4 tries which gives us 1-0.5787=0.4213.
My point is, don't focus on formulas. Think first and try to reduce a problem to something you already know. Analytical thinking > memory. That's kinda the way a mathematician is supposed to think anyway
Most of these small probability and statistics questions don't require any formal training
P.S. also if you happen to be familiar with simple combinatorics, most of the simple probability problems can be reduced to combinatorics problem: may not necessarily be as elegant but if you're familiar with that field better, it'll definitely be more efficient for you
I think in the paragraph about Q6 you meant to write 5/6 three times, but otherwise I agree.
I got them all. I tripped up on the Bayes Theorem problem at first, but then I got it.
Number 6 can be done another way. "Taking less than four tries" is equivalent to "not taking four or more tries". Taking four or more tries can be calculated easily because that just means that the first three attempts are failures, so that would be (5/6)^3. Then not taking four or more tries would be the compliment of that, or 1 - (5/6)^3. That's equivalent to the answer in the video.
Wow
Landed on the same answer through the same reasoning, too lazy to pull out a calculator
completely agree
I got all of them correct, the hardest was 9 because it took a while to enter all of the digits into my calculator
I just loaded up Desmos and used product formula (capital Pi) notation
Lol
7 was harder and 9 is fake actually, check people around you
Skill issue
1 - 365!/[(365-n)!*365^n], n=23
Another person mentioned it, but you should have both probabilites for false positive and negative for problem 10. They are not the same for almost every single real life application, and considering the difference between type 1 & 2 errors is crucial in statistics, which is the main application of probability theory!
Failed at level 3 😍 I'll update after studying some more and reattempting this
Im gonna love probabilty when I have it at 6th semester. Great video.
failed question 4 because I didn’t know what cards are in a deck
Same
Fr
for Q6 we can use the complement rule - meaning 1 Minus failure (which is 3 times of NOT getting a pair) - in this question (not all the time) this approach is easier than the geometric sequence one
9:30 We can also solve the question in this way.
P(getting a pair(doubles) in < 4 tries) = 1 - P(not getting a pair(doubles) in < 4 tries)
= 1 - Πi=1 to 3 P(not getting a pair(doubles) on try i)
= 1 - (30/36)³
≈ 0.4212.
For question 11, I just knew that one of the two questions would be in 1 of the 11 slots while the other is in one of the 10 remaining. Regardless of what slots the others are in.
That means there are 120 possibilities of slot combinations for those two questions.
I then calculated the number of possibilities that the questions are beside each other by considering number of permutations conditioning on the easiest question being in slot 1, then in slot 2, etc. and came up with 20. So it was a simple 20/110 or 2/11
Loved this vid. I got to level 10. At 11 i thought "pfft this is an easy one", made some really dumb mistake and failed.
Same. I just did 2x10/11!
I forgot the other numbers could also move
10 things you need to know about conic sections. 5 things you need to know about radical equations
Another solution method for 11 is that the easiest problem is either in position 1 or 11, in which case there is only one of 10 adjacent places, or in a position in between, in which case there are 2 of 10 adjacent places.
So the probability is 2/11*1/10 + 9/11*2/10 which is again 2/11.
for question 3, you just need to think that for any result the first dice gets (lets call it A), theres only 1 number in wich A+B=6, so you need to find a single number at the second dice, 1/6
Here is a question that I once solved which is very good.
Ques - Morse code uses dots and dashes, which are known to occur in the proportion of 3:4.
Suppose there is interference on the transmission line, and with probability 1/8 a dot is mistakenly received as a dash, and vice versa.
If we receive a dot, what is the probability that a dot was sent?
I'd say 3/8 times 7 or 21/8
We had it in seventh grade, I genuinely love math pls continue doing this videos.
These are 2 more interesting problems (but much harder than those mentioned in the video) for those who managed to solve all the problems provided and want a challenge! Great video btw!
Question 12: In the country of probaland, couples will have children until they have their first girl, at which point they will never have children again. On average, how many children does each couple have?
Question 13: In the country of probaland, couples will have children until they have their first girl, at which point they will never have children again. If you were to choose a random person in this land, what is the probability they are a boy?
Q12: 2
Q13:1/3
@@victory646813 is 0.5. Consider a fair game of coins (heads you win 1$, tails you lose 1$). There's clearly no strategy that gives a profit in the long run, thus no childbirth strategy can affect the overall ratio of boys to girls.
@@victory6468 😓could you explain me how you found that result please
Q12 : 2
Q13 : 1/2
@@massouleherrison3556 Use the concept of Expected probabilities
not difficult, but for Q10, the problem does not mention both true positive and true false have same conditional probability, we have to assume they are equal.
i think if you suppose the test 99.5% accurate, it works in both ways... No need to suppose that
@@fenrir6723 anyway, it is still an assumption, we should state that as it is a problem. For it is a practical case, feel free to assume what you like.
@fenrir6723 That is actually a very dangerous assumption and is most certainly not true for almost every real-life case. For pathegon detection tests(like COVID and the given problem), false positives are somewhat acceptable while false negatives can be devastating, so they are purposely built to have an extremely low false negative rate but not that low false positive rates. For fingerprint recognition, it's more dangerous to have a false positive so it's built the other way around.
@@fenrir6723I agree that the problem should've stated both probabilities, as that is needed info for problem solving and in my opinion is not a trait that can be just assumed to be true.
Its also irrational to think That 1 of 250 would be 0.44% chance... I believe he calculate it wrong
Thank you for this wonderful quiz
Thank you so much for the problem set!!! I was able to estimate where I stand. Would love to see similar quizzes for other concepts in mathematics.
Beginning of first year of studies in France (medecine). Got the 11 done
Thank you brother, informative and funny!! Great video, best wishes :)
I thoroughly enjoyed this video, and thanks for creating it! As for how I did, I was able to do #1 - #8, and #11. I did not try #10 because I don't like applying a theorem, in this case Bayes' Theorem, to a probability problem because it is usually a straightforward application.
The real story is about problem #9. Although I correctly approached it by trying to take the complement of the probability that no two students shared a birthday, as you did, I treated the numerator of this probability to be a combination problem, i.e., in how many different ways I can pick 23 days out of 365. Also, for the denominator, i.e., the total #possibilities, I simply took 365. As I took a snapshot peek at your solution to see how I faired, I realized that the numerator could be arrived at by treating it as a permutation problem, and I was way off with the denominator. Without playing your solution, I tried to arrive at your expression for the probability that no two students shared a birthday, and interestingly, arrived at it using an approach that was quite different from yours. So, for the numerator, i.e., #favorable events, your expression made me realize that this is a P (365, 23) permutation. For the denominator, my reasoning was as follows. The 1st student can have 365 possibilities for birthdays and for each of these possibilities, the 2nd student can have 365 possible birthdays, and for each of the 365 x 365 possibilities across the first two students the 3rd student can have 365 birthdays, and so on, to give 365^23 possibilities across 23 students. So, the probability that no two students shared a birthday comes to P (365, 23)/365^23, i.e., the same as your expression. However, what blew me away is that your approach was completely different, and a remarkable one, I must say. I also realized that your approach is akin to sampling with replacement. To understand this better, I personally reframed the problem as: What is the probability of picking the same student at least twice when randomly and sequentially picking 23 students from 365 students with replacement? So, the real kicker for me is that I took a peek at your expression and arrived at it in a way that is very different from yours! I thoroughly enjoyed this detour for problem #9 :)
for me, only the 9th one was challenging. discrete probability is generally easy. but when it gets continuous, then the trouble begins
They are standard (and kinda easy) questions in probability theory so i got all of them correctly. If i havent been taught probability theory tho i wouldnt score more than 3-4 of them right
In question number 11 28:55, the numerator should be 10! * 2 * 10, the extra 10 multiplication because the boxed question group could shift itself to 10 different places...
So there are 10 ways to arrange the group Q.1 and Q.11 across the blanks, 2! internal arrangements, and 10! Other element arrangements
I like your thinking but it would be 9! other element arrangements not 10!. Therefore what you are saying is that you would have 10*2*9! which is equal to 10! * 2 (which is how it is in the video)
Hope that helps!
I got tripped up for question 4 because I’m more familiar with logic, so when you said or I thought it meant it could the one of the two or both, but it turns out that the or you mentioned was what we call XOR (exclusive or), meaning that it can’t be both. In retrospect it was pretty obvious that you were talking about that 😅! Great video.
I thought it can be both?
The diagram shows that there are
- 10 hearts (no face)
- 9 faces (no heart)
- 3 heart and face
Total of 22
It's not an exclusive or. It means hearts and/or a face card. He subtracts cards with both so as to not double count them as both face and heart cards.
Matrices next?
We need one of these for integration
Q5 the answer is finding the probability of the “first two” cards of the deck are clubs, not what the question is asking
Got the first 6, not bad for being out of school over 2 years…
Bro an all-star now 🎉
For Q5 it can be simplified to 1/17
The qns were fairly ez if you've done highschool math/A level math. The only one i struggled with was the binomial one cause i forgot how they work, also the last one was definitely not the hardest.
I've only had calculus. This was a match.
Pls don’t stop making these videos
I love this channel
A different way of looking at that last one. There are 11 possible spots for the easiest question. In two of those (first and last) there is one of the remaining 10 spots where the hardest question can go and be next to it. In the nine other spots for the easiest question, the hard question can go in two of the ten remaining spots and be next to it. So 2/11 x 1/10 + 9/11 x 2/10 = .1818…
Alternatively, whenever the easiest and hardest question are next to each other they can be treated as a single object. Among 9 other questions there are 10! possible positionings, times 2 to account for the order of questions 1 and 11. Divide this by the total number of orderings (11!) for an answer of 2/11
14:31
For adults and children cant we do:
(6/10) * (5/9) * (4/8) * (3/7) = 0.0714
Order doesn't matter, even if we pick children first like:
(4/10) * (3/9) * (6/8) * (5/7)
Can someone explain whats wrong with this approach?
Can you please explain the numbers you picked and why you multiplied them?
the order matters because theres a different number of children and adults, since the denominator always decreases, it really does
The only thing you forget to account for is the position of the children and adults. Your solution assumes that the first 2 positions are ALWAYS children and the second 2 are ALWAYS adults.
But what if there's a child in the first and third position? That fits the criteria as well AND has the same probability as your case. There are 6 different degenerate such cases, ({1,2},{1,3},{1,4},{2,3},{2,4},{3,4}), hence you need to multiply your calculations by 6 to get the final answer.
You're solving this via permutation rule. Permutations assume order matters automatically, so you have to do every single ordering of the adults and kids. Or you can use combinations like he did and not worry about that.
drinking game: have a shot whenever “probability” is said 🎉
The First question is obviously 1/6 - you need 1 of the 6 sides. You DO NOT NEED the formula and if you do, you are Just making the solution redundantly complex and if used in algorithm, you would slow it Down by redundant calculation. Use LOGIC instead of patterns. - Tomas, senior software engineer
Bruh he just explained it in terms of math and not logic
You could have mentioned level 7 adheres to the "multivariate hypergeometric distribution"
question 6 is incredibly complicated
I am In Class XII and I was able to solve all 11 questions ❤ .. thanks for this video
These videos going to help me a lot in grade 10
Got through all of them :)
The last one was surprisingly easy (if you know factorials)
My answer for question 7 is 42.9%
I'm so good at math
EqualCount = 0
from random import randint
trials = 1_000_000
for trial in range(trials):
adults = 6
children = 4
for picks in range(4):
if randint(1, adults + children)
Are these questions good for highschool student's exam??? I know only one formula which was taught that is,
-->
probability of event happening - probability of event not happening = 1
Im happy that i got the first 5 questions
for question 5 why did you not divide the probability of A happening if it's in the equation?
never taken probability past fifth grade. let’s see how this goes
edit: i got 1, 2, 4, and part of 6 right 😭😭
Wouldn't OR exclude face cards that are hearts all together since they are both and therefore not either or the other?
Question 11: I would explain like this to younger student to avoid as much as possible factorial, but your explanation is great.
The pair easy/hard depends only on the placement of the easy question, which must always precede the hard one. There are 10 possible positions for the easy question (10 choose 1). Since the order can be reversed, the number of satisfying cases doubles.
The denominator is determined by the number of ways to arrange two questions within a set of 11 (number of arrangements of 2 elements within 11). The nine other questions do not interfere directly on the probability.
Therefore,
P = (2x10)/(11×10)=2/11
Even toughest question in this list is easier than easiest question of jee mains 😅😂, never mind
I got 1 wrong (question 7). Great test!
coins can flip tails heads and sideways
Mathematically 8 was harder than 9 which was harder than 10 which was harder than 11. however actually punching in the numbers to answer question 9 made me want to throw my calculator against the wall.
I just still dont feel like i get the concept behind question 6...Why is it that the odds of hitting the pair decreases as the tries increases, why isn't it increasing? Shouldnt the more you try, the higher your chances of success? please i really want to learn...
In that question if you want to get the pair in a specific turn you need to be unsuccessful the turns until that turn that's why the probability shrinks by time. I am not a native English speaker I hope you understand
P1 is the probability that you roll a pair, 1/6.
P2 is the probability that you rolled a pair (1/6) AND didn't roll a pair in the first one (5/6). Remember that we stop trying after we have rolled 1 pair.
Does this help you?
Q5 can be simplified to just 1/17
Sir please my humble request is to upload next vedio regarding
Intermediate value theorem ,
Mean value theorem, monotonicity, maxima and minima ,topic of Calculus
Yeah i started to lose points from q3 onwards. Im not good at probability
5th one is 1/17
Should have done 13/52 into 1/4
q11 was a bit of a bummer to be in level 11
Video on set theory
great video !!
Got all of them
I was supposed to get to the end… but I got stuck at 4 because I didn't know what a FACE CARD is.
Why didnt you simplify the answer in question 5 to 1/17?
Ok I'm at question 5 rn and WHY DID YOU NOT SIMPLIFY THAT FRACTION je devienne fou
THE LAST QUESTION!!!
WHY NOT SIMPLIFY THOSE FACTORIALS INTO 2/11?!?!? me angey it's 4am I should go to sleep
most of them are tough in
class 10 probability
I got 0.4% on Q.10.
They just say what is the chance of him getting the disease so it's still 0.4% regardless the test result
Not bad, although it's all at the level of intro level. I was expecting some complex multiple probability space problem
bro, actually the toughest was easier than few problems
Question 9 forgot about leap year
The last one was really easy idk
i thought he was going to bring out measure theoretic probabilities at level 10
I am so confused about number 9 right now. I figured I would try it and approached it by trying to count how many different ways there are for two students to share a birthday, which is quite simple, that being 22+21+20...+3+2+1, or 253. That represents the pairs of birthdays between those students. Each pair has a 1 in 365 chance to be the same. Basically you're just rolling two 365 sided dice 253 times and hope for at least one pair, or that's what I thought. But when calculating the probability (1-[364/365]²⁵³) my result was 50.0477154%, which is incredibly close to the answer you gave, but not quite there. But I double checked my inputs in the calculator multiple times and I can't find the flaw in my logic, yet it's still just ever so slightly off. Did I make a mistake? Did you make a mistake? Is it just complete coincidence? What's even more confounding to me is that if I do it to the power of 258 I'd get 50.728% (or 50.73% rounded) which is the result in the video. If you change the exponent by 1 the result is quite far away from that in comparison so it's not like you'd be expected to just get the answer eventually if you add or subtract 1 to/from the exponent enough. But 258 as an exponent makes absolutely no sense, so is it even more coincidence?
PDF of this video plsss it’s a humble request it will mean slot as it will help me in revision
Yo I feel like question 4 is the hardest bc the way you’ve worded it. It says heart OR face, not heart and/or face, meaning any cards that are in both categories are nullified. This means it would be 10(hearts that aren’t face) +9 (faces that aren’t hearts)/ 52(total cards). This should leave you with 19/52.
Ultimately language is vague, but generally in probability in English, we think of OR as being the inclusive OR, with exclusive OR (XOR) needing to be explicitly stated.
OR can be true if atleast one of them is True
AND can only be true if both statement is True
great vid
11th is easiest (string method)
Why was 11 the last one? I found it easier than all the others 4 and up.
The only one I couldn't get right was 8, yay
I think you got 6 wrong.
Imagine that after you threw the cube 4 times, and check if you got a double, in any of them. Getting a double in the first, does not negate your chance to get a double on the second
Your calculations showed the chance of getting a double once wheen throwing 4 times. Just to prove my point, you can calculate the opposite, which is (5/6)**4, and see that your answer plus it's opposite is less than 1
His calculations showed that P(1) + P(2) = 1/6 + 5/6 * 1/6.
This is the probability that the first or second roll is a double. We multiply the second roll by 5/6 bc that's the probability the first roll wasn't a double. We need that bc we only need 1 double before we stop rolling.
Also, you can calculate this using 1 - P(only rolling non-doubles 3 times), which is 1 - (5/6)**3. This will give the same outcome as in the video.
All done
Nice, sir. S Chitrai Kani
Hey bro can you make video about geometry cuz I lack on that I would like you to make video about it
anybody who competes in mathcounts or AMC can easily solve these 11 problems :P
Didnt divide by P(positive) on Q10 :/
I don’t know what is a deck of cards
my excuse if I don’t know the answer
Q7, isn't 0 even too?
This isn’t a true test lmao I’m a freshmen and got 1,2,3,7,8,11 correct. (Didn’t know 4,5 because of cards)
Could reach only till 9 level as I am in 11th grade when I will go upto 12 will again take it
got all of them ez
8/11
didn't get q9 right
got all of the other ones right
I could do the 7 but couldn't do the 6
No way i could solve the last, but coludnt solve number 5,6,7,8,9 and 10
I can solve 9 and 8 but solve 10 in easy way😂
how about derivites
Differentiation?
It doesnt get that hard, but its still a cool ideia
We already have a top 10 of Calculus tho
He did derivatives easiest to hardest a long time ago
Holly Molly just move on
wtf I solved 1 to 6 and then last one solved correctly