Question

Three coins have probabilities of head in a single toss as \(\frac{1}{4}\), \(\frac{1}{2}\), and \(\frac{3}{4}\) respectively. A player selects one coin at random and tosses it five times. The probability of obtaining two tails in five tosses is:

• A. \(\frac{85}{384}\)
• B. \(\frac{255}{384}\)
• C. \(\frac{125}{384}\)
• D. \(\frac{64}{384}\)

Hint:

When a coin is chosen randomly from multiple biased coins, use the law of total probability to average over all conditional probabilities.

Answer 1

The Correct Option is A

Step 1: Let the coins have head probabilities \(p_1 = \frac{1}{4}, p_2 = \frac{1}{2}, p_3 = \frac{3}{4}\).
Tail probabilities are \(q_1 = \frac{3}{4}, q_2 = \frac{1}{2}, q_3 = \frac{1}{4}\). Each coin is equally likely: \(P(C_i) = \frac{1}{3}\).
Step 2: Probability of exactly two tails in five tosses for each coin.
\[ P_i = \binom{5}{2} q_i^2 p_i^3 \] Compute each: \[ P_1 = 10 \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^3 = 10 \times \frac{9}{16} \times \frac{1}{64} = \frac{90}{1024} \] \[ P_2 = 10 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^3 = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{320}{1024} \] \[ P_3 = 10 \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^3 = 10 \times \frac{1}{16} \times \frac{27}{64} = \frac{270}{1024} \]
Step 3: Average over the three coins.
\[ P = \frac{1}{3}(P_1 + P_2 + P_3) = \frac{1}{3}\left(\frac{90 + 320 + 270}{1024}\right) = \frac{680}{3072} = \frac{85}{384} \] Final Answer: \[ \boxed{\frac{85}{384}} \]