Question

Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

• A. \(\frac{5}{8}\)
• B. \(\frac{1}{8}\)
• C. \(\frac{5}{16}\)
• D. 1

Hint:

For problems involving independent events, the probability of both events occurring is the product of their individual probabilities. When an outcome can be achieved in multiple mutually exclusive ways (e.g., same number of heads being 0, 1, 2, or 3), the total probability is the sum of the probabilities of each way.

Answer 1

The Correct Option is C

Step 1: Determine the probability distribution for one person tossing three coins.
Let X be the random variable representing the number of heads obtained when tossing three fair coins. The total number of possible outcomes is \(2^3 = 8\). The number of heads X can be 0, 1, 2, or 3. This follows a binomial distribution with n=3 and p=1/2.
- P(X=0) (TTT): \(\binom{3}{0}(\frac{1}{2})^3 = \frac{1}{8}\)
- P(X=1) (HTT, THT, TTH): \(\binom{3}{1}(\frac{1}{2})^3 = \frac{3}{8}\)
- P(X=2) (HHT, HTH, THH): \(\binom{3}{2}(\frac{1}{2})^3 = \frac{3}{8}\)
- P(X=3) (HHH): \(\binom{3}{3}(\frac{1}{2})^3 = \frac{1}{8}\)
Step 2: Calculate the probability of getting the same number of heads.
Let \(X_A\) be the number of heads for person A and \(X_B\) be the number of heads for person B. Since their tosses are independent, the probability that they both get k heads is \(P(X_A=k) \times P(X_B=k)\). We want to find the probability that \(X_A = X_B\). This can happen if both get 0 heads, or both get 1 head, or both get 2 heads, or both get 3 heads. \[ P(X_A = X_B) = P(X_A=0 \cap X_B=0) + P(X_A=1 \cap X_B=1) + P(X_A=2 \cap X_B=2) + P(X_A=3 \cap X_B=3) \] Due to independence: \[ P(X_A = X_B) = P(X_A=0)P(X_B=0) + P(X_A=1)P(X_B=1)\] \[+ P(X_A=2)P(X_B=2) + P(X_A=3)P(X_B=3) \] Step 3: Substitute the probabilities and compute the sum.
\[ P(X_A = X_B) = \left(\frac{1}{8}\right)\left(\frac{1}{8}\right) + \left(\frac{3}{8}\right)\left(\frac{3}{8}\right) + \left(\frac{3}{8}\right)\left(\frac{3}{8}\right) + \left(\frac{1}{8}\right)\left(\frac{1}{8}\right) \] \[ P(X_A = X_B) = \frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64} = \frac{1+9+9+1}{64} = \frac{20}{64} \] Simplifying the fraction: \[ P(X_A = X_B) = \frac{5 \times 4}{16 \times 4} = \frac{5}{16} \]