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I just saw the question in an exam paper. My friend and I have different answer and I am not sure which solution is correct, so I hope someone can help me.

There are 16 cups in a box, 11 are blue and 5 are white. Six cups are taken out from it at the same time. What is the probability of taking at least 4 white cups?

My solution:


My friend's solution:


My friend use a different approach and his answer is different from mine. I wonder if I am correct and why. Please show me the reason, thank you!

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Your method is correct. –  André Nicolas Apr 15 '13 at 8:02
@AndréNicolas Can you tell me why I am correct? After my friend told me his solution, I am so confused with the question ... –  redcap Apr 15 '13 at 8:22

1 Answer 1

up vote 1 down vote accepted

Your method is correct. There are $\binom{16}{6}$ equally likely ways to pick a collection of $6$ cups. (We can imagine the cups have secret ID numbers written in invisible ink.)

Your numerator counts the favourables. For example, the are $\binom{5}{4}$ ways to pick $4$ whites. For each of these ways there are $\binom{11}{2}$ ways to pick the accompanying blues, for a total of $\binom{5}{4}\binom{11}{2}$.

Your friend's proposed solution can be made to work, though it is perhaps a little trickier to do it correctly. The friend observed that counting order of picking, there are ${}_{16}P_6$ equally likely ways of picking $6$ cups. Now we count the number of favourables. Your friend's first term in the count of favourables is correct, the second is not.

There are $\binom{6}{2}$ ways of picking the locations of the $4$ blue cups. For each of these ways, the first chosen location can be filled in $11$ ways, and the second in $10$ ways. Then the remaining locations can be filled with white cups in ${}_5P_4$ ways. Multiply to get the total with $4$ white and $2$ blue. There are two mistakes in the proposed expression there. First, instead of $\binom{6}{2}$ there is ${}_6P_2$. Secondly, there is the $4!$ which may be a typo. For the placement of white cups in the $4$ available slots, either use the permutation symbol ${}_5P_4$, or decide the $4$ whites can be chosen in $\binom{5}{4}$ ways, and then permuted in $4!$ ways. That part turns out to be $(5)4!$, which is the same as $5!$.

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Thank you! I have asked my friend and he said 4! is a typo. (should be 5!) Thank you again for your explanation, I understand both of the two methods now :) –  redcap Apr 15 '13 at 9:17
You are welcome. I had figured the $4!$ was an unimportant error. But that still leaves an overcount in that part by a factor of $2$. That error is somewhat more structural. –  André Nicolas Apr 15 '13 at 9:30

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