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 of finding the expected value of the random variable \( X \), which represents the number of heads obtained when a biased coin is tossed 4 times, we can follow these steps:

### Step 1: Determine the probabilities of heads and tails
Let \( p \) be the probability of getting heads and \( q \) be the probability of getting tails. According to the problem, the probability of getting heads is twice that of tails. Thus, we can express this relationship as:
\[ p = 2q \]
Since the total probability must equal 1, we have:
\[ p + q = 1 \]
Substituting \( p \) in terms of \( q \):
\[ 2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3} \]
Now substituting back to find \( p \):
\[ p = 2q = 2 \times \frac{1}{3} = \frac{2}{3} \]

### Step 2: Define the random variable \( X \)
The random variable \( X \) can take values from 0 to 4 (the number of heads obtained in 4 tosses).

### Step 3: Calculate the probabilities for each value of \( X \)
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where \( n = 4 \) (number of tosses), \( p = \frac{2}{3} \) (probability of heads), and \( q = \frac{1}{3} \) (probability of tails).

Calculating for each value of \( X \):
- For \( X = 0 \):
\[ P(X = 0) = \binom{4}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^4 = 1 \cdot 1 \cdot \frac{1}{81} = \frac{1}{81} \]
- For \( X = 1 \):
\[ P(X = 1) = \binom{4}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^3 = 4 \cdot \frac{2}{3} \cdot \frac{1}{27} = \frac{8}{81} \]
- For \( X = 2 \):
\[ P(X = 2) = \binom{4}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^2 = 6 \cdot \frac{4}{9} \cdot \frac{1}{9} = \frac{24}{81} \]
- For \( X = 3 \):
\[ P(X = 3) = \binom{4}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^1 = 4 \cdot \frac{8}{27} \cdot \frac{1}{3} = \frac{32}{81} \]
- For \( X = 4 \):
\[ P(X = 4) = \binom{4}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^0 = 1 \cdot \frac{16}{81} \cdot 1 = \frac{16}{81} \]

### Step 4: Calculate the expected value \( E(X) \)
The expected value \( E(X) \) is calculated using the formula:
\[ E(X) = \sum_{k=0}^{n} k \cdot P(X = k) \]
Substituting the values:
\[ E(X) = 0 \cdot \frac{1}{81} + 1 \cdot \frac{8}{81} + 2 \cdot \frac{24}{81} + 3 \cdot \frac{32}{81} + 4 \cdot \frac{16}{81} \]
Calculating each term:
\[ E(X) = 0 + \frac{8}{81} + \frac{48}{81} + \frac{96}{81} + \frac{64}{81} = \frac{216}{81} = \frac{8}{3} \]

### Final Answer
Thus, the expected value of \( X \) is:
\[ \boxed{\frac{8}{3}} \]