Question

A coin is tossed three times. Let X denote the number of times a tail follows a head. If \(\mu\) and \(\sigma^2\) denote the mean and variance of X, then the value of \(64(\mu + \sigma^2)\) is:

• A. 51
• B. 48
• C. 32
• D. 64

Hint:

To calculate the expected value and variance for a probability distribution, use the formulas for mean and variance, and then apply them to the given probabilities.

Answer 1

The Correct Option is B

HHH \(\to 0\) 
HHT \(\to 0\) 
HTH \(\to 1\) 
HTT \(\to 0\) 
THH \(\to 1\) 
THT \(\to 1\) 
TTH \(\to 1\) 
TTT \(\to 0\) 
Probability distribution: \[ \mu = \sum x_i P_i = \frac{1}{2} \] \[ \sigma^2 = \sum x_i^2 P_i - \mu^2 = \frac{1}{2} \times 1^2 + \frac{1}{2} \times 1^2 - \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ 64(\mu + \sigma^2) = 64\left(\frac{1}{2} + \frac{1}{4}\right) = 64 \times \frac{3}{4} = 48 \]

Answer 2

Given: A coin is tossed three times. Let \( X \) = number of times a Tail follows a Head. We need to find \( 64(\mu + \sigma^2) \), where \( \mu = E(X) \) and \( \sigma^2 = Var(X) \).  

Step 1: List all possible outcomes Total outcomes = \( 2^3 = 8 \) | Outcome | X (Tail follows Head) | |----------|-----------------------| | HHH | 0 | | HHT | 1 | | HTH | 1 | | HTT | 1 | | THH | 0 | | THT | 1 | | TTH | 0 | | TTT | 0 | 
Step 2: Compute frequencies \( X = 0 \) occurs in 4 cases. \( X = 1 \) occurs in 4 cases. So, \( P(X=0) = \frac{4}{8} = \frac{1}{2} \) \( P(X=1) = \frac{4}{8} = \frac{1}{2} \) 
Step 3: Mean \[ \mu = E(X) = 0\cdot\frac{1}{2} + 1\cdot\frac{1}{2} = \frac{1}{2} \] 
Step 4: Variance \[ E(X^2) = 0^2\cdot\frac{1}{2} + 1^2\cdot\frac{1}{2} = \frac{1}{2} \] \[ \sigma^2 = E(X^2) - [E(X)]^2 = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] 
Step 5: Compute required value \[ 64(\mu + \sigma^2) = 64\left(\frac{1}{2} + \frac{1}{4}\right) = 64\left(\frac{3}{4}\right) = 48 \] 
∴ Correct option: 2) 48