Probability of Specific Head Outcomes in Coin Tosses

To find the value of $96p$, we first calculate the probability $p$ of the given compound event occurring when a fair coin is tossed 8 times.

Let $X_i \in \{0, 1\}$ denote the outcome of the $i$-th toss, where $X_i = 1$ represents Heads and $X_i = 0$ represents Tails, for $i \in \{1, 2, \dots, 8\}$.

The total number of possible outcomes for 8 tosses is: $\text{Total outcomes} = 2^8 = 256$

We are given two conditions:

1. Exactly 4 heads appear in the first 6 tosses: $\sum_{i=1}^6 X_i = (X_1 + X_2 + X_3) + (X_4 + X_5 + X_6) = 4$
2. Exactly 3 heads appear in the last 5 tosses: $\sum_{i=4}^8 X_i = (X_4 + X_5 + X_6) + (X_7 + X_8) = 3$

Notice that the two sets of tosses overlap at tosses 4, 5, and 6. Let $k = X_4 + X_5 + X_6$ be the total number of heads in these three overlapping tosses. Since there are 3 tosses, $k \in \{0, 1, 2, 3\}$.

Using $k$, we can rewrite the two conditions:

• $X_1 + X_2 + X_3 = 4 - k$
• $X_7 + X_8 = 3 - k$

Since $X_1 + X_2 + X_3 \le 3$ and $X_7 + X_8 \ge 0$, the possible values for $k$ must satisfy: $4 - k \le 3 \implies k \ge 1$ $3 - k \ge 0 \implies k \le 3$

Thus, $k$ can take values $1, 2,$ or $3$. We analyze each case separately:

1. Case $k = 1$:

   • Number of ways to choose 1 head in tosses $\{4, 5, 6\}$: $\binom{3}{1} = 3$
   • Number of ways to choose $4 - 1 = 3$ heads in tosses $\{1, 2, 3\}$: $\binom{3}{3} = 1$
   • Number of ways to choose $3 - 1 = 2$ heads in tosses $\{7, 8\}$: $\binom{2}{2} = 1$ $\text{Favorable outcomes for } k=1 = 3 \times 1 \times 1 = 3$

2. Case $k = 2$:

   • Number of ways to choose 2 heads in tosses $\{4, 5, 6\}$: $\binom{3}{2} = 3$
   • Number of ways to choose $4 - 2 = 2$ heads in tosses $\{1, 2, 3\}$: $\binom{3}{2} = 3$
   • Number of ways to choose $3 - 2 = 1$ head in tosses $\{7, 8\}$: $\binom{2}{1} = 2$ $\text{Favorable outcomes for } k=2 = 3 \times 3 \times 2 = 18$

3. Case $k = 3$:

   • Number of ways to choose 3 heads in tosses $\{4, 5, 6\}$: $\binom{3}{3} = 1$
   • Number of ways to choose $4 - 3 = 1$ head in tosses $\{1, 2, 3\}$: $\binom{3}{1} = 3$
   • Number of ways to choose $3 - 3 = 0$ heads in tosses $\{7, 8\}$: $\binom{2}{0} = 1$ $\text{Favorable outcomes for } k=3 = 1 \times 3 \times 1 = 3$

Summing the favorable outcomes from all cases gives: $\text{Total favorable outcomes} = 3 + 18 + 3 = 24$

Thus, the probability $p$ is: $p = \frac{24}{256} = \frac{3}{32}$

We are required to find $96p$: $96p = 96 \times \frac{3}{32} = 3 \times 3 = 9$