Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is

A biased coin is tossed 5 times. Let $$p$$ = probability of heads.

Use the given condition P(4 heads) = P(5 heads).

$$\binom{5}{4}p^4(1-p)^1 = \binom{5}{5}p^5(1-p)^0$$

$$5p^4(1-p) = p^5$$

$$5(1-p) = p$$

$$5 - 5p = p$$

$$p = \frac{5}{6}, \quad q = 1 - p = \frac{1}{6}$$

Find P(at most 2 heads).

$$P(X \leq 2) = P(0) + P(1) + P(2)$$

$$P(0) = \binom{5}{0}\left(\frac{5}{6}\right)^0\left(\frac{1}{6}\right)^5 = \frac{1}{6^5}$$

$$P(1) = \binom{5}{1}\left(\frac{5}{6}\right)^1\left(\frac{1}{6}\right)^4 = \frac{25}{6^5}$$

$$P(2) = \binom{5}{2}\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)^3 = 10 \cdot \frac{25}{6^5} = \frac{250}{6^5}$$

$$P(X \leq 2) = \frac{1 + 25 + 250}{6^5} = \frac{276}{6^5} = \frac{276}{6 \cdot 6^4} = \frac{46}{6^4}$$

The correct answer is Option A: $$\frac{46}{6^4}$$.