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