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

Let's solve the problem by understanding the scenario and calculating the probability that both A and B get the same number of heads when they each toss three fair coins independently.

First, understand the possible outcomes when one person tosses 3 coins: 

Number of Heads	Possible Combinations	Probability
0 Heads	TTT	\(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)
1 Head	TTH, THT, HTT	\(\binom{3}{1} \left(\frac{1}{2}\right)^3 = \frac{3}{8}\)
2 Heads	THH, HTH, HHT	\(\binom{3}{2} \left(\frac{1}{2}\right)^3 = \frac{3}{8}\)
3 Heads	HHH	\(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)

Now, both A and B must get the same number of heads. Let's compute the probability for each case:

• Both get 0 Heads: \(\frac{1}{8} \times \frac{1}{8} = \frac{1}{64}\)
• Both get 1 Head: \(\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}\)
• Both get 2 Heads: \(\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}\)
• Both get 3 Heads: \(\frac{1}{8} \times \frac{1}{8} = \frac{1}{64}\)

Add these probabilities together to find the total probability that A and B get the same number of heads:

\(\frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64} = \frac{20}{64} = \frac{5}{16}\)

Thus, the probability that both A and B get the same number of heads is \(\frac{5}{16}\), which corresponds to option C.