Question

A hunter knows that a deer is hidden in one of the two near by bushes, the probability of its being hidden in bush-I being $\frac{4}{5}$. The hunter having a rifle containing 10 bullets decides to fire them all at bush-I or II. It is known that each shot may hit one of the two bushes, independently of the other with probability $\frac{1}{2}$. Number of bullets must he fire on bush-I to hit the animal with maximum probability is (Assume that the bullet hitting the bush also hits the animal).

Answer 1

6

Answer 2

We first note that the deer is in bush‑I with probability

$$P_I=\frac{4}{5}$$

and in bush‑II with probability

$$P_{II}=\frac{1}{5}.$$

If the hunter fires $x$ bullets at bush‑I and $(10-x)$ bullets at bush‑II, then each bullet (independently) hits its chosen bush with probability $\frac{1}{2}$. Hence, the probability that at least one bullet hits bush‑I is

$$1-\left(\frac{1}{2}\right)^x,$$

and similarly, the probability that at least one bullet hits bush‑II is

$$1-\left(\frac{1}{2}\right)^{10-x}.$$

Thus, the overall probability $P$ of hitting the deer is

$$P(x)=\frac{4}{5}\left[1-\left(\frac{1}{2}\right)^x\right] + \frac{1}{5}\left[1-\left(\frac{1}{2}\right)^{10-x}\right].$$

Since the overall constant “1” does not affect maximization, we need to minimize

$$Q(x)=\frac{4}{5}\left(\frac{1}{2}\right)^x+\frac{1}{5}\left(\frac{1}{2}\right)^{10-x}.$$

To find the optimal $x$, we consider $x$ as a continuous variable and differentiate $Q(x)$ with respect to $x$.

Express the terms using exponentials:

$$\left(\frac{1}{2}\right)^x = e^{-x\ln 2} \quad \text{and} \quad \left(\frac{1}{2}\right)^{10-x} = e^{-(10-x)\ln 2} = e^{-10\ln2}e^{x\ln 2}.$$

Thus,

$$Q(x)=\frac{4}{5}e^{-x\ln2} + \frac{1}{5}e^{-10\ln2}e^{x\ln2}.$$

Differentiate:

$$\frac{dQ}{dx} = \frac{4}{5}(-\ln2)e^{-x\ln2} + \frac{1}{5}e^{-10\ln2}(\ln2)e^{x\ln2}.$$

Set $\frac{dQ}{dx}=0$:

$$-\frac{4}{5}\ln2\,e^{-x\ln2} + \frac{1}{5}\ln2\,e^{-10\ln2}e^{x\ln2}=0.$$

Cancel $\frac{1}{5}\ln2$ (which is nonzero) to obtain:

$$-4\,e^{-x\ln2} + e^{-10\ln2}e^{x\ln2}=0.$$

Rearrange:

$$e^{-10\ln2}e^{x\ln2} = 4\, e^{-x\ln2}.$$

Divide both sides by $e^{-x\ln2}$:

$$e^{-10\ln2}e^{2x\ln2} = 4.$$

This is equivalent to:

$$e^{(2x-10)\ln2} = 4 \quad \Rightarrow \quad 2^{2x-10} = 4.$$

Since $4=2^2$, we have:

$$2x-10=2 \quad \Rightarrow \quad 2x=12 \quad \Rightarrow \quad x=6.$$

Thus, the hunter should fire 6 bullets at bush‑I (and the remaining 4 at bush‑II) to maximize the probability of hitting the deer.