Two players toss 4 coins each. The probability that they both obtain the same number of heads is `5//256` b. `1//16` c. `35//128` d. none of these

Options

A. `5//256`
B. `1//16`
C. `35//128`
D. None of these

Explanation

To solve the problem of finding the probability that two players tossing 4 coins each obtain the same number of heads, we can follow these steps:

### Step 1: Understand the Problem
Each player tosses 4 coins, and we need to find the probability that both players end up with the same number of heads.

### Step 2: Possible Outcomes
The possible outcomes for the number of heads (H) each player can get when tossing 4 coins are 0, 1, 2, 3, or 4 heads.

### Step 3: Calculate Individual Probabilities
Using the binomial probability formula:
\[ P(r) = \binom{n}{r} p^r q^{n-r} \]
where:
- \( n \) = number of trials (4 coins)
- \( r \) = number of successes (number of heads)
- \( p \) = probability of success (getting heads = \( \frac{1}{2} \))
- \( q \) = probability of failure (getting tails = \( \frac{1}{2} \))

For our case:
- \( n = 4 \)
- \( p = \frac{1}{2} \)
- \( q = \frac{1}{2} \)

### Step 4: Calculate Probabilities for Each Case
We will calculate the probability for each case (0 heads, 1 head, 2 heads, 3 heads, and 4 heads) and then square it (since both players must achieve the same number of heads).

1. **Probability of 0 heads:**
\[ P(0) = \binom{4}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 = 1 \cdot 1 \cdot \frac{1}{16} = \frac{1}{16} \]
So, \( P(0, 0) = \left(\frac{1}{16}\right)^2 = \frac{1}{256} \)

2. **Probability of 1 head:**
\[ P(1) = \binom{4}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^3 = 4 \cdot \frac{1}{2} \cdot \frac{1}{8} = \frac{4}{16} = \frac{1}{4} \]
So, \( P(1, 1) = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)

3. **Probability of 2 heads:**
\[ P(2) = \binom{4}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = 6 \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{6}{16} = \frac{3}{8} \]
So, \( P(2, 2) = \left(\frac{3}{8}\right)^2 = \frac{9}{64} \)

4. **Probability of 3 heads:**
\[ P(3) = \binom{4}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^1 = 4 \cdot \frac{1}{8} \cdot \frac{1}{2} = \frac{4}{16} = \frac{1}{4} \]
So, \( P(3, 3) = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)

5. **Probability of 4 heads:**
\[ P(4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^0 = 1 \cdot \frac{1}{16} \cdot 1 = \frac{1}{16} \]
So, \( P(4, 4) = \left(\frac{1}{16}\right)^2 = \frac{1}{256} \)

### Step 5: Total Probability
Now, we sum the probabilities of both players getting the same number of heads:
\[ P(\text{same heads}) = P(0, 0) + P(1, 1) + P(2, 2) + P(3, 3) + P(4, 4) \]
\[ = \frac{1}{256} + \frac{1}{16} + \frac{9}{64} + \frac{1}{16} + \frac{1}{256} \]
Converting all terms to a common denominator (256):
\[ = \frac{1}{256} + \frac{16}{256} + \frac{36}{256} + \frac{16}{256} + \frac{1}{256} = \frac{1 + 16 + 36 + 16 + 1}{256} = \frac{70}{256} \]

### Step 6: Simplifying the Result
Now, we simplify \( \frac{70}{256} \):
\[ \frac{70}{256} = \frac{35}{128} \]

### Final Answer
Thus, the probability that both players obtain the same number of heads is:
\[ \boxed{\frac{35}{128}} \]