A coin is tossed 8 times. If the probability that exactly 4 heads...

Let the outcomes of the $8$ tosses be denoted by $X_1, X_2, X_3, X_4, X_5, X_6, X_7, X_8$.

Let $A$ be the number of heads in the first $3$ tosses ($X_1, X_2, X_3$).
Let $B$ be the number of heads in the next $3$ tosses ($X_4, X_5, X_6$).
Let $C$ be the number of heads in the last $2$ tosses ($X_7, X_8$).

According to the given conditions:
Number of heads in the first six tosses: $A + B = 4$
Number of heads in the last five tosses: $B + C = 3$

Since $C$ is the number of heads in $2$ tosses, $C \le 2$. Thus, $B = 3 - C \ge 1$. Also, $B \le 3$.
The possible values for $B$ are $1, 2, 3$. We analyze each case:

Case 1: $B = 1$
Then $C = 2$ and $A = 3$.
Number of ways = $^{3}C_{3} \times ^{3}C_{1} \times ^{2}C_{2} = 1 \times 3 \times 1 = 3$

Case 2: $B = 2$
Then $C = 1$ and $A = 2$.
Number of ways = $^{3}C_{2} \times ^{3}C_{2} \times ^{2}C_{1} = 3 \times 3 \times 2 = 18$

Case 3: $B = 3$
Then $C = 0$ and $A = 1$.
Number of ways = $^{3}C_{1} \times ^{3}C_{3} \times ^{2}C_{0} = 3 \times 1 \times 1 = 3$

Total number of favorable outcomes = $3 + 18 + 3 = 24$
Total number of possible outcomes for $8$ coin tosses = $2^8 = 256$

The probability $p$ is given by:
$p = \dfrac{24}{256} = \dfrac{3}{32}$

We need to find the value of $96p$:
$96p = 96 \times \dfrac{3}{32} = 3 \times 3 = 9$

Answer: $9$