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 find the probability that a man gets a head in a coin toss before he gets a 5 or 6 in a dice throw, we can break down the problem 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 tail (T) when tossing the coin is \( P(T) = \frac{1}{2} \).
- The probability of getting a 5 or 6 when throwing the die is \( P(5 \text{ or } 6) = \frac{2}{6} = \frac{1}{3} \).
- The probability of not getting a 5 or 6 (getting 1, 2, 3, or 4) when throwing the die is \( P(\text{not } 5 \text{ or } 6) = \frac{4}{6} = \frac{2}{3} \).

### Step 2: Analyze the first toss
The man starts by tossing the coin:
- If he gets a head (H) on the first toss, the game ends, and he wins. The probability of this event is \( \frac{1}{2} \).

### Step 3: Analyze the case of getting a tail
- If he gets a tail (T) on the first toss (which happens with probability \( \frac{1}{2} \)), he then rolls the die:
- If he rolls a 5 or 6 (with probability \( \frac{1}{3} \)), he loses.
- If he rolls a number that is not 5 or 6 (with probability \( \frac{2}{3} \)), he goes back to tossing the coin again.

### Step 4: Set up the equation
Let \( P \) be the probability that he gets a head before getting a 5 or 6. We can express \( P \) as follows:

\[ P = P(H) + P(T) \times P(\text{not } 5 \text{ or } 6) \times P \]

Substituting the probabilities we have:

\[ P = \frac{1}{2} + \frac{1}{2} \times \frac{2}{3} \times P \]

### Step 5: Simplify the equation
This simplifies to:

\[ P = \frac{1}{2} + \frac{1}{3} P \]

Now, we can isolate \( P \):

\[ P - \frac{1}{3} P = \frac{1}{2} \]

\[ \frac{2}{3} P = \frac{1}{2} \]

### Step 6: Solve for \( P \)
Multiplying both sides by \( \frac{3}{2} \):

\[ P = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} \]

### Conclusion
Thus, the probability that he gets a head in the coin toss before he gets a 5 or 6 in the dice throw is \( \frac{3}{4} \).