A man alternately tosses a coin and throws a die beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is `3//4` b. `1//2` c. `1//3` d. none of these

Options

A. `3//4`
B. `1//2`
C. `1//3`
D. None of these

Explanation

To solve the problem, we need to determine the probability that a man gets a head when tossing a coin before he rolls a 5 or 6 on a die. Let's break it down step by step.

### Step 1: Define the probabilities
- The probability of getting a head (H) when tossing the coin is:
\[ P(H) = \frac{1}{2} \]
- The probability of getting a 5 or 6 (F) when rolling the die is:
\[ P(F) = \frac{2}{6} = \frac{1}{3} \]

### Step 2: Define the complementary probabilities
- The probability of not getting a head (T) when tossing the coin is:
\[ P(T) = 1 - P(H) = 1 - \frac{1}{2} = \frac{1}{2} \]
- The probability of not getting a 5 or 6 when rolling the die is:
\[ P(\text{not } F) = 1 - P(F) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Calculate the overall probability
The man alternates between tossing the coin and rolling the die. The probability that he gets a head before rolling a 5 or 6 can be expressed as follows:

1. He can get a head on the first toss:
\[ P(H) = \frac{1}{2} \]

2. If he does not get a head, he must also not roll a 5 or 6. The probability of this happening is:
\[ P(T) \times P(\text{not } F) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \]
After this, he is back to the original situation, where he can again toss the coin.

Thus, the total probability \( P \) can be expressed as:
\[ P = P(H) + P(T) \times P(\text{not } F) \times P \]
Substituting the known values:
\[ P = \frac{1}{2} + \frac{1}{3} P \]

### Step 4: Solve for P
Rearranging the equation gives:
\[ P - \frac{1}{3} P = \frac{1}{2} \]
\[ \frac{2}{3} P = \frac{1}{2} \]
Multiplying both sides by \( \frac{3}{2} \):
\[ P = \frac{3}{4} \]

### Conclusion
The probability that he gets a head before rolling a 5 or 6 is:
\[ \boxed{\frac{3}{4}} \]