To solve the problem, we need to find the probability \( p \) that two consecutive heads occur on the fourth and fifth toss of a fair coin. Let's break down the steps:
### Step 1: Understanding the conditions
We need to determine the outcomes of the first five tosses of the coin such that:
- The fourth toss is a head (H).
- The fifth toss is also a head (H).
- The third toss must be a tail (T) to avoid having two consecutive heads before the fourth toss.
Thus, the sequence of the first five tosses must look like this:
- \( X_1, X_2, T, H, H \)
where \( X_1 \) and \( X_2 \) can be either heads (H) or tails (T), but cannot both be heads.
### Step 2: Analyzing possible outcomes for \( X_1 \) and \( X_2 \)
Since we need to avoid having two consecutive heads in the first three tosses, we can have the following combinations for \( X_1 \) and \( X_2 \):
1. \( (H, T) \)
2. \( (T, H) \)
3. \( (T, T) \)
### Step 3: Calculating probabilities for each case
1. **Case 1: \( (H, T) \)**
- Probability = \( P(H) \times P(T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
2. **Case 2: \( (T, H) \)**
- Probability = \( P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
3. **Case 3: \( (T, T) \)**
- Probability = \( P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
### Step 4: Total probability for \( X_1 \) and \( X_2 \)
Now, we sum the probabilities of the valid cases:
- Total probability \( P_1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \)
### Step 5: Probability of the fixed tosses
The probability of the fixed tosses (T, H, H) is:
- \( P(T) \times P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)
### Step 6: Total probability \( p \)
Thus, the total probability \( p \) that two consecutive heads occur on the fourth and fifth toss is:
\[
p = P_1 \times P(T, H, H) = \frac{3}{4} \times \frac{1}{8} = \frac{3}{32}
\]
### Step 7: Finding \( \frac{30}{p} \)
Now, we need to find \( \frac{30}{p} \):
\[
\frac{30}{p} = \frac{30}{\frac{3}{32}} = 30 \times \frac{32}{3} = \frac{960}{3} = 320
\]
### Final Answer
Thus, the value of \( \frac{30}{p} \) is \( \boxed{320} \).
