In a box containing 36 strawberries, 2 of them are rotten. Kyle randomly picked 5 out of these strawberries.

a) What is the probability of having at least 1 rotten strawberry among the five?

b) How many strawberries should be picked so that the probability of having exactly 2 rotten strawberries among them equals $\frac{2}{35}$

$\begin{pmatrix} 36 \\ 5\end{pmatrix} = \frac{36!}{5!31!} = 376,992$ ways to pick 5 strawberries.

$\begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix} = 2$ different ways that rotten strawberries can be picked

$\begin{pmatrix} 34 \\ 3 \end{pmatrix} + \begin{pmatrix} 34 \\ 4 \end{pmatrix} = 52,360$ ways 3 non rotten strawberries can be picked

Then I combined all of these to get

$$\frac{2 * 52,360}{376,992} = .2778$$

So I concluded that you have a 27.78% chance of picking at least 1 rotten strawberry among the five. However the answer says $\frac{11}{42} = .2619 = 26.19\%$. I am not confident that I set the problem up correctly.

For $b)$, I have no idea how to even approach thinking about the problem...

  • 1
    $\begingroup$ Why do you add in your approach to the first part? But here is a different approach: the complimentary event is "all five strawberries are good", which has probability $\binom {34}5\big /\binom {36}5$. $\endgroup$ – lulu Sep 14 '19 at 0:19
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    $\begingroup$ For the second part, the probability of having both bad strawberries, if you choose $k$, is $\binom {34}{k-2}\big / \binom {36}k$. $\endgroup$ – lulu Sep 14 '19 at 0:22
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    $\begingroup$ Ah, I see what you were trying. To do it your way, you have to handle two cases. You get exactly one bad one or you get both. The number of ways to get exactly one bad one is $\binom 21 \times \binom {34}{4}$. The number of ways to get both bad ones is $\binom 22\times \binom {34}3$. Now you can add. $\endgroup$ – lulu Sep 14 '19 at 0:34
  • $\begingroup$ @lulu why is it k-2? $\endgroup$ – Evan Kim Sep 14 '19 at 5:01
  • $\begingroup$ Because you used $2$ of the slots for the rotten ones. $\endgroup$ – lulu Sep 14 '19 at 12:11

Think of $X$ as a random variable where $X$ is the number of rotten strawberries picked in $n$ trials. So $X$ is a hypergeometric random variable and $$ P(X=k)=\dfrac{{2\choose k} {34 \choose n-k}}{36 \choose n} \tag{1} $$ We want $P(X=2)=\dfrac{2}{35}$ so just let $k=2$ in (1) and solve for $n$.

  • $\begingroup$ hmm, I see what you are doing, but we have technically not learned about random variables yet (it is in the next chapter) so I am not sure if this is the way they were looking for us to solve this problem by $\endgroup$ – Evan Kim Sep 14 '19 at 15:44
  • $\begingroup$ I think your book is just trying to motivate the material that is going to be presented in the next chapter. So the author wants you to "come up" with the idea of a random variable on your own since that is how you would solve this problem. $\endgroup$ – alpastor Sep 14 '19 at 15:46

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