Question

A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is \(p\), then \(96p\) is equal to ____.

Answer 1

Correct Answer: 35

Step 1: Understanding the Concept:
We are given a sequence of 8 coin tosses.
The condition specifies the number of heads in two overlapping groups of tosses.
Group 1 consists of the first 6 tosses (\(T_1\) to \(T_6\)), which must contain exactly 4 heads.
Group 2 consists of the last 5 tosses (\(T_4\) to \(T_8\)), which must contain exactly 3 heads.
The overlap between these two groups is the set of tosses \(\{T_4, T_5, T_6\}\).
Step 2: Key Formula or Approach:
Let the number of heads in the overlapping tosses \(\{T_4, T_5, T_6\}\) be \(k\).
Since there are 3 tosses in the overlap, \(k\) can range from 0 to 3.
Let \(H_1\) be the number of heads in the first 3 tosses (\(T_1, T_2, T_3\)).
Let \(H_2\) be the number of heads in the last 2 tosses (\(T_7, T_8\)).
We can set up the equations based on the given conditions:
\(H_1 + k = 4\) (from the first 6 tosses).
\(k + H_2 = 3\) (from the last 5 tosses).
The number of ways to achieve this for a specific \(k\) is \(\binom{3}{H_1} \times \binom{3}{k} \times \binom{2}{H_2}\).
Step 3: Detailed Explanation:
From the equations, we have \(H_1 = 4 - k\) and \(H_2 = 3 - k\).
Since \(0 \le H_1 \le 3\), we get \(0 \le 4 - k \le 3 \implies 1 \le k \le 4\).
Since \(0 \le H_2 \le 2\), we get \(0 \le 3 - k \le 2 \implies 1 \le k \le 3\).
Taking the intersection of these conditions, \(k\) can only be 1, 2, or 3.
Now we calculate the number of favorable outcomes for each valid \(k\):
Case 1 (\(k = 1\)): \(H_1 = 3\), \(H_2 = 2\).
Ways = \(\binom{3}{3} \times \binom{3}{1} \times \binom{2}{2} = 1 \times 3 \times 1 = 3\).
Case 2 (\(k = 2\)): \(H_1 = 2\), \(H_2 = 1\).
Ways = \(\binom{3}{2} \times \binom{3}{2} \times \binom{2}{1} = 3 \times 3 \times 2 = 18\).
Case 3 (\(k = 3\)): \(H_1 = 1\), \(H_2 = 0\).
Ways = \(\binom{3}{1} \times \binom{3}{3} \times \binom{2}{0} = 3 \times 1 \times 1 = 3\).
Total favorable outcomes = \(3 + 18 + 3 = 24\).
The total number of possible outcomes for 8 coin tosses is \(2^8 = 256\).
Thus, the probability \(p = \frac{24}{256} = \frac{3}{32}\).
Step 4: Final Answer:
We need to find the value of \(96p\):
\(96p = 96 \times \frac{3}{32} = 3 \times 3 = 9\).