Question

Player \( P_1 \) tosses 4 fair coins and player \( P_2 \) tosses a fair die independently of \( P_1 \). The probability that the number of heads observed is more than the number on the upper face of the die, equals

• A. \( \frac{7}{16} \)
• B. \( \frac{5}{32} \)
• C. \( \frac{17}{96} \)
• D. \( \frac{21}{64} \)

Hint:

For probability involving independent events, calculate the joint probability by multiplying the individual probabilities for each event.

Answer 1

The Correct Option is C

Step 1: Understand the events.
Player \( P_1 \) tosses 4 fair coins, so the possible number of heads is 0, 1, 2, 3, or 4. Player \( P_2 \) rolls a fair die, so the possible outcomes for the die are 1, 2, 3, 4, 5, and 6.
Step 2: Calculate the probability.
The probability of getting \( k \) heads for \( P_1 \) is given by the binomial distribution: \[ P(\text{k heads}) = \binom{4}{k} \left( \frac{1}{2} \right)^4 \] The probability that \( P_1 \)'s number of heads is greater than \( P_2 \)'s die roll is calculated by summing the appropriate probabilities: \[ P(\text{k heads}>\text{die roll}) \]
Step 3: Final calculation.
The total probability is \( \frac{17}{96} \), which is the correct answer (C).