A man alternately tosses a coin and throw a dice, beginning with the coin. The probability that he gets a head in coin before he gets a 5 or 6 in dice, is

Options

A. `(3)/(4)`
B. `(1)/(2)`
C. `(1)/(3)`
D. `(2)/(3)`

Answer

To solve the problem of finding the probability that a man gets a head in a coin toss before he gets a 5 or 6 on a dice, we can break down the events and calculate the probabilities step by step.

### Step 1: Define the Events
Let:
- Event A: Getting a head on the coin toss.
- Event B: Getting a 5 or 6 on the dice.

### Step 2: Calculate Individual Probabilities
- The probability of getting a head (P(A)) when tossing the coin is:
\[ P(A) = \frac{1}{2} \]
- The probability of getting a 5 or 6 (P(B)) when throwing the dice is:
\[ P(B) = \frac{2}{6} = \frac{1}{3} \]
- The probability of not getting a head (P(A')) is:
\[ P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \]
- The probability of not getting a 5 or 6 (P(B')) is:
\[ P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Analyze the Sequence of Events
The man alternates between tossing the coin and throwing the dice. The process continues until he either gets a head or rolls a 5 or 6. The possible scenarios can be broken down as follows:

1. He gets a head on the first toss (probability = \( \frac{1}{2} \)).
2. He gets a tail on the first toss and does not roll a 5 or 6 on the dice. Then he gets another chance to toss the coin (probability = \( \frac{1}{2} \times \frac{2}{3} \)).
3. This process continues, forming a geometric series.

### Step 4: Set Up the Probability Expression
The total probability (P) that he gets a head before rolling a 5 or 6 can be expressed as:
\[ P = P(A) + P(A') \cdot P(B') \cdot P \]
Substituting the probabilities:
\[ P = \frac{1}{2} + \left(\frac{1}{2} \cdot \frac{2}{3}\right) P \]
This simplifies to:
\[ P = \frac{1}{2} + \frac{1}{3} P \]

### Step 5: Solve for P
Rearranging the equation:
\[ P - \frac{1}{3} P = \frac{1}{2} \]
\[ \frac{2}{3} P = \frac{1}{2} \]
Multiplying both sides by \( \frac{3}{2} \):
\[ P = \frac{3}{4} \]

### Conclusion
Thus, the probability that he gets a head in the coin toss before he gets a 5 or 6 in the dice is:
\[ \boxed{\frac{3}{4}} \]