A biased coin is tossed repeatedly until a tail appears for the first time. The head is 3 times likely to appear as the tail. The probability that the number of tosses required will be more than 4, given that in first two toss no tail has occur, is

Options

A. `(3)/(4)`
B. `(7)/(8)`
C. `(9)/(16)`
D. `(11)/(32)`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Define the probabilities
Let \( P_H \) be the probability of getting heads and \( P_T \) be the probability of getting tails. According to the problem, heads are 3 times as likely to occur as tails. Therefore, we can write:

\[ P_H = 3P_T \]

Since the total probability must equal 1, we have:

\[ P_H + P_T = 1 \]

### Step 2: Substitute and solve for probabilities
Substituting \( P_H \) in the total probability equation:

\[ 3P_T + P_T = 1 \]

This simplifies to:

\[ 4P_T = 1 \]

Thus, we find:

\[ P_T = \frac{1}{4} \]

Now substituting back to find \( P_H \):

\[ P_H = 3P_T = 3 \times \frac{1}{4} = \frac{3}{4} \]

### Step 3: Given condition
We are given that no tails have occurred in the first two tosses. This means the first two tosses must be heads. The probability of getting heads in the first two tosses is:

\[ P(HH) = P_H \times P_H = \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) = \frac{9}{16} \]

However, since we are given that this event has occurred, we treat it as a certainty (probability = 1).

### Step 4: Find the probability of more than 4 tosses
We need to find the probability that the number of tosses required will be more than 4, given that the first two tosses were heads. We denote this event as \( P(X > 4 | HH) \).

Using the complement rule:

\[ P(X > 4 | HH) = 1 - P(X \leq 4 | HH) \]

### Step 5: Calculate \( P(X \leq 4 | HH) \)
We can express this as:

\[ P(X \leq 4 | HH) = P(X = 3 | HH) + P(X = 4 | HH) \]

**For \( P(X = 3 | HH) \):**
This means we have heads on the first two tosses and tails on the third toss:

\[ P(X = 3 | HH) = P(HH) \times P(T) = 1 \times P_T = \frac{1}{4} \]

**For \( P(X = 4 | HH) \):**
This means we have heads on the first three tosses and tails on the fourth toss:

\[ P(X = 4 | HH) = P(HH) \times P(H) \times P(T) = 1 \times P_H \times P_T = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16} \]

### Step 6: Combine the probabilities
Now we can combine these probabilities:

\[ P(X \leq 4 | HH) = P(X = 3 | HH) + P(X = 4 | HH) = \frac{1}{4} + \frac{3}{16} \]

To add these fractions, we convert \( \frac{1}{4} \) to sixteenths:

\[ \frac{1}{4} = \frac{4}{16} \]

Thus:

\[ P(X \leq 4 | HH) = \frac{4}{16} + \frac{3}{16} = \frac{7}{16} \]

### Step 7: Final calculation
Now substituting back into our complement:

\[ P(X > 4 | HH) = 1 - P(X \leq 4 | HH) = 1 - \frac{7}{16} = \frac{9}{16} \]

### Final Answer
The probability that the number of tosses required will be more than 4, given that in the first two tosses no tail has occurred, is:

\[ \boxed{\frac{9}{16}} \]