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 a biased coin is tossed 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 \( q \). Since the probability of heads is twice that of tails, we have:
\[ p = 2q \]
Given that the total probability must equal 1, we can write:
\[ p + q = 1 \]
Substituting \( p \):
\[ 2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3} \]
Therefore,
\[ p = 2q = 2 \times \frac{1}{3} = \frac{2}{3} \]

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

3. **Define the random variable**:
The random variable \( X \) can take values \( 0, 1, 2, 3, \) or \( 4 \) (the number of heads).

4. **Calculate the expected value \( E(X) \)**:
The expected value \( E(X) \) can be calculated using the formula:
\[ E(X) = \sum_{r=0}^{n} r \cdot P(X = r) \]
where \( P(X = r) \) is given by the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

5. **Calculate \( P(X = r) \) for \( r = 0, 1, 2, 3, 4 \)**:
- For \( r = 0 \):
\[ P(X = 0) = \binom{4}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^4 = 1 \cdot 1 \cdot \frac{1}{81} = \frac{1}{81} \]
- For \( r = 1 \):
\[ P(X = 1) = \binom{4}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^3 = 4 \cdot \frac{2}{3} \cdot \frac{1}{27} = \frac{8}{81} \]
- For \( r = 2 \):
\[ P(X = 2) = \binom{4}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^2 = 6 \cdot \frac{4}{9} \cdot \frac{1}{9} = \frac{24}{81} \]
- For \( r = 3 \):
\[ P(X = 3) = \binom{4}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^1 = 4 \cdot \frac{8}{27} \cdot \frac{1}{3} = \frac{32}{81} \]
- For \( r = 4 \):
\[ P(X = 4) = \binom{4}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^0 = 1 \cdot \frac{16}{81} \cdot 1 = \frac{16}{81} \]

6. **Calculate \( E(X) \)**:
Now we can substitute the values into the expected value formula:
\[ E(X) = 0 \cdot \frac{1}{81} + 1 \cdot \frac{8}{81} + 2 \cdot \frac{24}{81} + 3 \cdot \frac{32}{81} + 4 \cdot \frac{16}{81} \]
\[ = 0 + \frac{8}{81} + \frac{48}{81} + \frac{96}{81} + \frac{64}{81} \]
\[ = \frac{8 + 48 + 96 + 64}{81} = \frac{216}{81} = \frac{8}{3} \]

### Final Answer:
The expected value of \( X \) is \( \frac{8}{3} \).