from what I understood in the class, Greg solved the question assuming at least two people picked their own colors card and showed the probability as 1/3. Can someone explain why that’s the case?
it can also be 4/5 * 3/4 * 2/3 * 1/2 * 1 * 5p2 which is not 1/3 (if the first 3 people are picking different colored cards)
So, my question is why are we assuming the people getting the same color are drawing the cards first?
My answer was not very clear, I will try to explain in detail how I solved it.
let’s start with your scenario: w-w-w-r-r
w means wrong (different color)
r means right (same color)
the probability of this scenario is not 4/5 * 3/4 * 2/3 * 1/2 * 1 because you did not consider that the first three players must not choose the card of the last two players(otherwise your scenario will be impossible). the only way for your scenario to exist is that the first three players interchange their cards (for example the first choose the second card, the second choose the third and the third choose the first card). In other words, It’s equivalent to letting the last two players choose the cards first. this scenario is equivalent to any other scenario including s-s-w-w-w.
finally the probability = (1/5)(1/4)(2/3)(1/2)(1)x 5!/2!3! = 1/6.
I see, the error was 5C2 instead of 5P2 wich must result in 1/6 instrad of 1/3
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Can you explain your solution thoroughly please? Still cannot understand how is the first case as
4/5*3/4*2/3*1/2*1?
This is wrong as I mentioned in my detailed analysis.
Okay, I got it but finally the probability (1/5)(1/4)(2/3)(1/2)(1)x 5!/2!3! = 1/6. In this portion why are multiplying it by 2/3. Shouldn’t it be 1/3?
okay, I think I got it thanks.
Welcome
because the 3rd person needs to pick a card out of the 3 that isnt his color. So, 2 of the 3 cards
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Got it Thanks!
by any chance do you have the screeshot of the 29.18 question from todays class?
I don’t have the question, but I do remember the question. If x/y=25.18, what is the sum of the largest and smallest of first 10 possible remainders of x/y?
I hope this helps.
yeah that helps, thanks alot.
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Hi everyone,
So today Greg, gave us this problem in class which is mentioned above. I have been trying to solve this since morning, as I am quite confused with the to approach Greg took here. Can someone help me out with this? I will be highly obliged!
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See the member area, the answer in the class was incorrect.
Edit: here’s how I would do it. Let the people be A, B, C, D and E. Now, how many ways can we pick two out of these five (that is, the ones that would get the original colour back)? There are 5 ways to pick the first member, and 4 ways to pick the second member. Hence the number of ways is 5 \times 4 = 20. But then exactly half of them would be duplicates (for instance, the pair (A, B) and (B, A) are the same), hence we divide by 2. This gets us 10 unique ways.
Now, without loss of generality, let A and B be the ones we know will get the right pair. Then the other three people must not get the correct pair. How many ways can we arrange three people in such a way that none of them get the right pair? That’s 2; namely the mapping (C, D, E) := (D, E, C) and (E, C, D). Hence the total number of ways is 10 \times 2 = 20. As there are 5! = 120 ways of arranging five unique colours to five people, the probability is
\frac{20}{120} = \frac{1}{6}
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