A biased coin with probability of getting head is twice that of tail, is tossed 4 times If a random variable X is number of heads obtained, then expected value of X is :

Options

A. ` ( 2 ) /(3 ) `
B. ` (8 ) /(3) `
C. `2`
D. ` (16 ) /(3) `

Explanation

To solve the problem, we need to find the expected value of the random variable \( X \), which represents the number of heads obtained when tossing a biased coin 4 times. The probability of getting heads is twice that of getting tails.

### Step-by-Step Solution:

1. **Define the probabilities**:
Let the probability of getting tails be \( P(T) = q \). Since the probability of getting heads is twice that of tails, we have:
\[ P(H) = 2q \]
The total probability must equal 1:
\[ P(H) + P(T) = 1 \implies 2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3} \]
Therefore, the probability of getting heads is:
\[ P(H) = 2q = 2 \times \frac{1}{3} = \frac{2}{3} \]

2. **Identify the number of trials**:
The coin is tossed \( n = 4 \) times.

3. **Calculate the expected value**:
The expected value \( E(X) \) for a binomial distribution can be calculated using the formula:
\[ E(X) = n \cdot P(H) \]
Substituting the values we have:
\[ E(X) = 4 \cdot \frac{2}{3} = \frac{8}{3} \]

### Final Result:
The expected value of \( X \) is:
\[ E(X) = \frac{8}{3} \]