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In a group of $10$ people, $60\%$ have brown eyes. Two people are to be selected at random from the group. What is the probability that neither person selected will have brown eyes?

How do I do this problem? $6$ people have brown eyes and $4$ people don't.

The possibility of people not have brown eyes is:

$$4 * 3 * 2 * 1 = 24$$

What to do?

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3 Answers 3

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Hint: two different people are selected from the group of ten. In how many ways can you select those two people from the four who do not have brown eyes? In other words, suppose the group is labeled

$$\{B1, B2, B3, B4, B5, B6, N1, N2, N3, N4\}.$$ Then how many ways can you choose two different people from the $N$ subgroup?

Next, how many ways can you choose two different people from the entire group?

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  • $\begingroup$ Well @heropup, the first way is 4 ways, then the second is 3 ways? So: 4*3 = 12 ways? $\endgroup$
    – Ama
    Mar 1, 2015 at 10:25
  • $\begingroup$ Then how should I determine the probability? $\endgroup$
    – Ama
    Mar 1, 2015 at 10:26
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On the first pick, there is $\frac4{10}$ chance that the person does not have brown eyes. On the second pick, after picking a person without brown eyes, there is $\frac39$ chance that the person does not have brown eyes. $$ \frac4{10}\cdot\frac39=\frac2{15} $$

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EDIT: I was overcounting in my previous solution.

Probability of an event = Number of favourable outcomes / Total number of outcomes

Here total outcomes = Number of ways to choose 2 people out of 10, which is $10C2$

Number of favourable outcomes = Number of ways to choose one non brown eyed person * Number of ways to choose another non - brown eyed person which is $(\binom 4 1 \binom 3 1)/2$

We are dividing it by two because here, the order of selection does not matter.

So our final probability is $(\binom 4 1 \binom 3 1)/(2 * \binom {10} 2)$

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    $\begingroup$ Be careful. If Amy has green eyes and Barbara has blue eyes, then selecting Amy first and Barbara second yields the same result as selecting Barbara first and Amy second. You have overcounted. $\endgroup$
    – N. F. Taussig
    Mar 1, 2015 at 11:58
  • $\begingroup$ @N. F. Taussig Thanks for pointing it out. I will edit my answer. $\endgroup$
    – Very Professional Username
    Mar 1, 2015 at 12:24
  • $\begingroup$ Your answer is now correct. Note that since the order of selection does not matter, the number of ways of selecting two people who do not have brown eyes is $$\binom{4}{2} = \frac{1}{2!}\binom{4}{1}\binom{3}{1} = \frac{1}{2}\binom{4}{1}\binom{3}{1}$$ $\endgroup$
    – N. F. Taussig
    Mar 1, 2015 at 12:41

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