A fair coin is tossed repeatedly until two consecutive heads are obtained. If the probability that 2 consecutive heads occur on fourth and fifth toss is p, then `(30)/(p)` is equal to

Answer

To solve the problem, we need to find the probability \( p \) that two consecutive heads occur on the fourth and fifth tosses of a fair coin. Let's break down the steps to find this probability.

### Step 1: Understand the sequence of tosses
We need to analyze the outcomes of the first five tosses. The requirement is that the fourth and fifth tosses must be heads (HH), and the third toss must be a tail (T) to ensure that we do not have two consecutive heads before the fourth toss.

### Step 2: Fix the outcomes for the third, fourth, and fifth tosses
From the requirement, we can fix the outcomes as follows:
- Toss 3: T (Tail)
- Toss 4: H (Head)
- Toss 5: H (Head)

This gives us the sequence: _ _ T H H, where the first two tosses can be either heads or tails.

### Step 3: Determine the possible outcomes for the first two tosses
The first two tosses can be:
1. HH
2. HT
3. TH
4. TT

### Step 4: Calculate the probability for each case
Since the coin is fair, the probability of each toss (Head or Tail) is \( \frac{1}{2} \).

- The probability of each of the first two tosses is \( \frac{1}{2} \) for each toss.
- Therefore, the probability for the first two tosses is:
\[ P(\text{first two tosses}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \]

### Step 5: Combine the probabilities
Now, we combine the fixed outcomes for tosses 3, 4, and 5:
- The probability of T on toss 3 is \( \frac{1}{2} \).
- The probability of H on toss 4 is \( \frac{1}{2} \).
- The probability of H on toss 5 is \( \frac{1}{2} \).

Thus, the total probability \( p \) is:
\[ p = P(\text{first two}) \times P(T) \times P(H) \times P(H) = \frac{1}{4} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \times \frac{1}{8} = \frac{1}{32} \]

### Step 6: Calculate \( \frac{30}{p} \)
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 \( 320 \).

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