What is the probability that at least $11$ boxes must be purchased in order to obtain two of the prizes?

Question

The probability that a randomly selected box of a certain type of cereal has a particular prize is $0.17$. Suppose that you purchase box after box until you have obtained $2$ of these prizes.
(a) What is the probability that you purchase exactly 7 boxes?
(b) What is the probability that you purchase at least 11 boxes?

I got the answer for a) which is $0.0683$ (using the binomial distribution: ($6C5 \cdot 0.17^2 \cdot 0.83^5$), but I cannot get the answer for b).

I tried using the summation of the same binomial distribution from $9$ to infinity but I can't get an exact number for the summation. I also tried finding each probability from $2$ (minimum) to $8$ (maximum to exclude $9$), adding them up and then subtracting the number from $1$ but am unable to figure it out.

I keep seeing "use negative binomial distribution" but I don't know what that means. If you could write the steps to get the answer that would help a lot.

Answer 1

You solved the first problem correctly.

Notice that you must purchase at least $11$ boxes if you obtain at most one prize in the first $10$ purchases. Therefore, the probability that you purchase at least $11$ boxes is $$\Pr(X \leq 1) = \Pr(X = 0) + \Pr(X = 1) = \binom{10}{0}(0.17)^{0}(0.83)^{10} + \binom{10}{1}(0.17)^{1}(0.83)^9$$

Answer 2

The negative binomial distribution is a distribution using to find the number of trials until a success. This is what you used in part a actually.

For part b, we can reword the question from purchasing at least 11 boxes to having less than 2 prizes in the first 10 boxes (Any case where this is not true you would have stopped purchasing boxes before your 11th purchase). To compute this, we can sum binomial probabilities of 0 successes and 1 success in 10 trials:

${10 \choose 0}\times0.17^0\times0.83^{10} + {10 \choose 1}\times0.17^1\times0.83^{9} = 0.473$