If `Aa n dB` each toss three coins. The probability that both get the same number of heads is `1//9` b. `3//16` c. `5//16` d. `3//8`

Options

A. `1//9`
B. `3//16`
C. `5//16`
D. `3//8`

Explanation

To solve the problem of finding the probability that both A and B get the same number of heads when each tosses three coins, we can follow these steps:

### Step 1: Understand the scenario
Each player (A and B) tosses three coins. The possible outcomes for the number of heads (H) can be 0, 1, 2, or 3.

### Step 2: Calculate the probability for one player
The probability of getting exactly r heads when tossing n coins can be calculated using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
where:
- \(n\) = number of trials (coins tossed) = 3
- \(p\) = probability of getting heads = \(\frac{1}{2}\)
- \(1-p\) = probability of getting tails = \(\frac{1}{2}\)

### Step 3: Calculate probabilities for each possible outcome
We will calculate the probabilities for A getting 0, 1, 2, and 3 heads.

1. **For 0 heads (r = 0)**:
\[ P(A = 0) = \binom{3}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^3 = 1 \cdot 1 \cdot \frac{1}{8} = \frac{1}{8} \]

2. **For 1 head (r = 1)**:
\[ P(A = 1) = \binom{3}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = 3 \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{3}{8} \]

3. **For 2 heads (r = 2)**:
\[ P(A = 2) = \binom{3}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^1 = 3 \cdot \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{8} \]

4. **For 3 heads (r = 3)**:
\[ P(A = 3) = \binom{3}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^0 = 1 \cdot \frac{1}{8} \cdot 1 = \frac{1}{8} \]

### Step 4: Combine probabilities for both players
Since A and B are independent, the probability that both A and B get the same number of heads is the sum of the probabilities of each case where they get the same number of heads:

\[ P(\text{same heads}) = P(A = 0)P(B = 0) + P(A = 1)P(B = 1) + P(A = 2)P(B = 2) + P(A = 3)P(B = 3) \]

Calculating this:
1. For 0 heads:
\[ P(A = 0)P(B = 0) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64} \]

2. For 1 head:
\[ P(A = 1)P(B = 1) = \frac{3}{8} \cdot \frac{3}{8} = \frac{9}{64} \]

3. For 2 heads:
\[ P(A = 2)P(B = 2) = \frac{3}{8} \cdot \frac{3}{8} = \frac{9}{64} \]

4. For 3 heads:
\[ P(A = 3)P(B = 3) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64} \]

Now, summing these probabilities:
\[ P(\text{same heads}) = \frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64} = \frac{20}{64} = \frac{5}{16} \]

### Final Answer
Thus, the probability that both A and B get the same number of heads is:
\[ \frac{5}{16} \]