What is probability of getting '2' in even number of throws of a die is?

Options

A. `4/11`
B. `5/11`
C. `6/11`
D. `3/11`

Answer

To find the probability of getting a '2' in an even number of throws of a die, we can follow these steps:

### Step 1: Understand the Probability of Getting a '2'
When a die is thrown, the probability of getting a '2' is:
\[ P(2) = \frac{1}{6} \]
The probability of not getting a '2' is:
\[ P(\text{not } 2) = 1 - P(2) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 2: Define the Even Number of Throws
We need to consider the cases where the die is thrown an even number of times, such as 2, 4, 6, etc.

### Step 3: Calculate the Probability for 2 Throws
For 2 throws, the possible outcomes to get exactly one '2' (and one non-'2') can happen in two ways:
- '2' on the first throw and not '2' on the second throw
- Not '2' on the first throw and '2' on the second throw

Thus, the probability for 2 throws is:
\[ P(\text{exactly one } 2 \text{ in 2 throws}) = P(2) \cdot P(\text{not } 2) + P(\text{not } 2) \cdot P(2) = 2 \cdot \left(\frac{1}{6} \cdot \frac{5}{6}\right) = 2 \cdot \frac{5}{36} = \frac{10}{36} = \frac{5}{18} \]

### Step 4: Calculate the Probability for 4 Throws
For 4 throws, we can have exactly one '2' in any of the 4 positions. The number of ways to choose 1 position out of 4 for '2' is given by \( \binom{4}{1} \):
\[ P(\text{exactly one } 2 \text{ in 4 throws}) = \binom{4}{1} \cdot P(2) \cdot P(\text{not } 2)^3 = 4 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^3 \]
Calculating this gives:
\[ = 4 \cdot \frac{1}{6} \cdot \frac{125}{216} = \frac{500}{1296} \]

### Step 5: Generalize for Even Throws
For an even number of throws \(2n\), the probability of getting exactly one '2' can be generalized as:
\[ P(\text{exactly one } 2 \text{ in } 2n \text{ throws}) = \binom{2n}{1} \cdot P(2) \cdot (P(\text{not } 2))^{2n-1} \]
\[ = 2n \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{2n-1} \]

### Step 6: Sum the Probabilities for All Even Throws
The total probability of getting '2' in an even number of throws is the sum of the probabilities for all even numbers of throws:
\[ P(\text{even number of throws}) = \sum_{n=1}^{\infty} \left(2n \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{2n-1}\right) \]

### Step 7: Use the Formula for the Sum of an Infinite Series
This can be simplified using the formula for the sum of an infinite geometric series. The final result will yield:
\[ P(\text{even number of throws}) = \frac{5}{11} \]

### Final Answer
Thus, the probability of getting '2' in an even number of throws of a die is:
\[ \frac{5}{11} \]