People are late 23% of the time. Out of 7 people, what is the probability that exactly 5 are late.

Question

A doctor knows 23 % of all her patients are late for their appointments. Given 7 randomly selected patients, what is the approximate probability that exactly 5 of them are late for their appointments?

Here is my approach:

P = $(0.23)^5*(1-0.23)^2$

Is this the right idea?

Answer 1

Almost, but not quite.

One question to consider is, which $5$ of the $7$ people are late?

There are $21$ different combinations of $5$ people that can be picked from a pool of $7$ people.

There is a $(0.23)^5*(1-0.23)^2$ chance that each particular combination is late, for a total probability of $21(0.23)^5*(1-0.23)^2$.

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Another question you may consider is, what's the probability you flip three coins and get $2$ heads?

The answer isn't $(0.5)^2(1-0.5)$, as you may easily see. It is actually $3/8$, as there are $3$ ways to choose $2$ heads from $3$ possible coins.

Answer 2

You need a binomial coefficient of $\binom{7}{5}$ in front of your answer.