A fair coin is tossed two times independently and X denotes the number of heads. Then a fair 6-faced die is thrown at random (independently of the tosses) and Y denotes the number on the top of the die. What is the probability that the value of X + Y is 4?

Correct option is D

Given:

• A fair coin is tossed two times independently
• Let X = number of heads → Possible values of X: 0, 1, 2
• A fair 6-faced die is rolled → Let Y = number on top → Y ranges from 1 to 6
• We are to find the probability that X + Y = 4

Solution:

Step 1: Probabilities for X (number of heads)

Sample space for 2 coin tosses:

• HH → X = 2
• HT, TH → X = 1
• TT → X = 0

So:

• P(X = 0) = 1/4
• P(X = 1) = 2/4 = 1/2
• P(X = 2) = 1/4

Step 2: Find all combinations where X + Y = 4

We want values of X and Y such that X + Y = 4

Check all possible X values:

​$$\begin{aligned}&\text{If } X = 0 \Rightarrow Y = 4 \Rightarrow P = P(X = 0) \cdot P(Y = 4) = \frac{1}{4} \cdot \frac{1}{6} \\&\text{If } X = 1 \Rightarrow Y = 3 \Rightarrow P = \frac{1}{2} \cdot \frac{1}{6} \\&\text{If } X = 2 \Rightarrow Y = 2 \Rightarrow P = \frac{1}{4} \cdot \frac{1}{6} \\[10pt]&\text{Total probability:} \\&P(X + Y = 4) = \frac{1}{4} \cdot \frac{1}{6} + \frac{1}{2} \cdot \frac{1}{6} + \frac{1}{4} \cdot \frac{1}{6} = \frac{1 + 2 + 1}{24} = \frac{4}{24} = \frac{1}{6}\end{aligned}$$​​

Correct Option: (D)$$\frac{1}{6}$$​​