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I was reviewing for the SAT II Subject Test (Mathematics 2). On a practice test, I came across the following question:

Twenty-five percent of a group of unrelated students are only children. The students are asked one at a time whether they are only children. What is the probability that the 5th student asked is the first only child?

Using the binomial probability theorem, $P=\binom{n}{r}p^{r}q^{n-r}$, I substituted in the values to get $P=\binom{5}{1}(\frac{1}{4})^{1}(\frac{3}{4})^{4}$. I used my calculator to resolve this expression to $P\approx 0.40$. However, that was not even one of the choices; the book says the correct answer is $P\approx 0.08$. What did I do wrong here?

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up vote 6 down vote accepted

Let's simplify the discussion a bit by agreeing that each child is asked "Are you an only child?" and they answer "Yes" or "No".

To say that the fifth child asked is the first "Yes" is precisely the same as to say that the fifth is a "Yes" and the first four children are "No"s.

$$\begin{align*} P(\text{5}^{\text{th}}\text{ child is the first Yes}) &= P(\text{first four children are No's})\times P(\text{5}^{\text{th}}\text{ child is a Yes})\\ &=P(\text{a child is a No})^4\times P(\text{a child is a Yes})\\ &=(0.75)^4\times(0.25)\\ &=0.0791015625\\ &\approx 0.08 \end{align*}$$

In terms of what you did wrong: when you use a formula, you need to ask yourself what the formula computes, and figure out if that's the thing the question is asking for. The formula $$\binom{n}{k}p^k q^{n-k}$$ gives you the probability that $k$ successes occur in $n$ trials, regardless of what order those successes or failures occur in. So, when you wrote $$\binom{5}{1}(0.25)^1(0.75)^4$$ you computed the probability that, when five children are asked the question, exactly one of them says "Yes", whereas the probability that exactly one of them says "Yes" and that child is the 5th child is just $$(0.25)^1(0.75)^4.$$

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Thank you for your answer. You were very thorough. – adamada May 25 '14 at 2:50

You are using the incorrect distribution. The binomial distribution refers to the probability of observing some number of successes of a fixed number $n$ of independent and identically distributed trials. Moreover, the binomial distribution is not concerned with the order in which successes are observed. Therefore, it is not an appropriate probability model for this question.

The appropriate model is geometric, which calculates the probability of the first success occurring after a particular number of trials. That is to say, the question asks for the probability that the first success is observed on the fifth trial, implying that four failures must be observed, then one success. This probability is simply $(3/4)^4 (1/4)$, due to the assumption of independence of trials. But notice that it is not $\dbinom{5}{1} (1/4)(3/4)^4$, because the binomial probability counts the outcomes where, for instance, the first student confirms they are an only child, and the next four students say they are not. Since there are $\binom{5}{1}$ such ways of choosing the one student who is an only child out of the five, we can directly see how the binomial distribution does not apply here.

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Thank you for your answer. – adamada May 25 '14 at 2:49

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