Maximum Probability to hit the bear. A bear hides itself either behind a bush $a$ with the probability $\frac{9}{25}$ or behind bush $b$ with probability $\frac{16}{25}$ . A hunter has $5$ bullets each of which can be fired either at bush $a$ or $b$ . Hunter hits each target independently with an accuracy of $\frac{1}{4}$ . How many bullets can be fired at the bush to hit the bear with maximum probability (assume that the bullet hitting the bush also hit the bear)?
How to incorporate the term maximum probability in the solution?

The hunter should pursue sequential strategy and not decide on distribution of shots in advance. Initial odds of the bear being behind bush $b$ are $r_0=\frac {16}9$ and it's clear that the first shot shout be at $b$. If he misses, odds get multiplied by $\frac 3 4$ to yield $r_1=r_0\cdot \frac 3 4=\frac 4 3$ so on the second shot he should shoot at $b$ as well. If he misses again, $r_2=1$ and then he is indifferent, so, for simplicity, he chooses to shoot at $b$ again. If he misses, he'll shoot at $a$ and if he misses then, he'll be indifferent.
Hence optimal strategies are of the form $bbbax$ and $bbabx$ where $x\in\{a,b\}$.

If the hunter doesn't know the outcome of his shooting (the bear is hiding and not roaring as I initially assumed), then we need to minimize $$\frac 9{25}(\frac 3 4)^i+\frac {16}{25}(\frac 3 4)^{5-i}$$ for $i$- number of shots fired at bush $a$. Minimum is achieved for $i=1,2$ leading to a similar set of strategies but without minimization of the shots fired.
