In a throw of a dice the probability of getting one is even number of throws is

Options

A. `(5)/(36)`
B. `(5)/(11)`
C. `(6)/(11)`
D. `1/6`

Answer

To find the probability of getting a 1 in an even number of throws of a die, we can break down the problem step by step.

### Step 1: Understand the Problem
We need to find the probability of rolling a 1 on a die in an even number of throws (2, 4, 6, ...). The probability of rolling a 1 on a single throw of a die is \( P(1) = \frac{1}{6} \), and the probability of not rolling a 1 is \( P(\text{not 1}) = \frac{5}{6} \).

### Step 2: Calculate Probability for Each Even Throw
1. **For 2 throws**:
- The first throw must not be a 1, and the second throw must be a 1.
- Probability:
\[ P(2) = P(\text{not 1}) \times P(1) = \left(\frac{5}{6}\right) \times \left(\frac{1}{6}\right) = \frac{5}{36} \]

2. **For 4 throws**:
- The first three throws must not be a 1, and the fourth throw must be a 1.
- Probability:
\[ P(4) = P(\text{not 1})^3 \times P(1) = \left(\frac{5}{6}\right)^3 \times \left(\frac{1}{6}\right) = \frac{125}{1296} \]

3. **For 6 throws**:
- The first five throws must not be a 1, and the sixth throw must be a 1.
- Probability:
\[ P(6) = P(\text{not 1})^5 \times P(1) = \left(\frac{5}{6}\right)^5 \times \left(\frac{1}{6}\right) = \frac{3125}{7776} \]

### Step 3: Generalize the Probability for Even Throws
For an even number of throws \( 2n \):
\[ P(2n) = \left(\frac{5}{6}\right)^{2n-1} \times \left(\frac{1}{6}\right) \]

### Step 4: Sum the Probabilities for All Even Throws
We need to sum the probabilities for \( n = 1, 2, 3, \ldots \):
\[ P(\text{even}) = P(2) + P(4) + P(6) + \ldots \]
This can be expressed as an infinite series:
\[ P(\text{even}) = \frac{1}{6} \left( \left(\frac{5}{6}\right) + \left(\frac{5}{6}\right)^3 + \left(\frac{5}{6}\right)^5 + \ldots \right) \]

### Step 5: Recognize the Series as a Geometric Series
The series inside the parentheses is a geometric series where:
- First term \( a = \frac{5}{6} \)
- Common ratio \( r = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \)

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Thus,
\[ S = \frac{\frac{5}{6}}{1 - \frac{25}{36}} = \frac{\frac{5}{6}}{\frac{11}{36}} = \frac{5 \times 36}{6 \times 11} = \frac{30}{11} \]

### Step 6: Calculate the Total Probability
Now substituting back into the equation for \( P(\text{even}) \):
\[ P(\text{even}) = \frac{1}{6} \times \frac{30}{11} = \frac{5}{11} \]

### Final Answer
The probability of getting a 1 in an even number of throws of a die is \( \frac{5}{11} \).

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