# Probability: Taking Cups From A Box

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:

$$P=\frac{\binom54\binom{11}2+\binom55\binom{11}1}{\binom{16}6}$$

My friend's solution:

$$P=\frac{5!\times11\times6+5!\times11\times10\times_6\text{P}_2}{_{16}\text{P}_6}$$

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!

-
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

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 $5!$.
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