# challenging probability question

I got this from a friend.

There are $43$ potential essay prompts. Six will show up on the history final, and you will have to choose $3$ to write about. What is the minimum number essay prompts that you should prepare for, such that the probability that the final has more than $3$ prompts that you haven't prepared for is less than $20\%$?

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There are altogether $\binom{43}6$ possible sets of six prompts that could show up on the final. Suppose that you prepare for $p$ of the prompts, leaving $43-p$ ‘dangerous’ prompts.

• There are $\displaystyle\binom{43-p}4\binom{p}2$ ways to choose $4$ dangerous and $2$ safe prompts.
• There are $\displaystyle\binom{43-p}5\binom{p}1$ ways to choose $5$ dangerous and $1$ safe prompt.
• There are $\displaystyle\binom{43-p}6\binom{p}0$ ways to choose $6$ dangerous and $0$ safe prompts.

Thus, the probability of getting a dangerous exam is

$$\frac{\dbinom{43-p}4\dbinom{p}2+\dbinom{43-p}5p+\dbinom{43-p}6}{\dbinom{43}6}\;.\tag{1}$$

I don’t immediately see a nice way to solve for the minimum $p$ for which $(1)$ is less than $\frac15$, but actual computation shows that it’s $p=25$:

$$\frac{\dbinom{18}4\dbinom{25}2+\dbinom{18}525+\dbinom{18}6}{\dbinom{43}6}=\frac{1,150,764}{6,096,454}\approx0.18876\;,$$ and

$$\frac{\dbinom{19}4\dbinom{24}2+\dbinom{19}524+\dbinom{19}6}{\dbinom{43}6}=\frac{1,375,980}{6,096,454}\approx0.22570\;.$$

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