If eight coins are tossed together, then the probability of getting exactly 3 heads is

Options

A. `1/256`
B. `7/32`
C. `5/32`
D. `3/32`

Explanation

To find the probability of getting exactly 3 heads when tossing 8 coins, we can use the binomial probability formula. The steps are as follows:

### Step-by-Step Solution:

1. **Identify the parameters:**
- Number of trials (n): 8 (since we are tossing 8 coins)
- Number of successes (x): 3 (we want exactly 3 heads)
- Probability of success (p): 1/2 (the probability of getting a head)
- Probability of failure (q): 1/2 (the probability of getting a tail)

2. **Use the binomial probability formula:**
The probability of getting exactly x successes in n trials is given by:
\[ P(X = x) = \binom{n}{x} p^x q^{n-x} \]
Here, \(\binom{n}{x}\) is the binomial coefficient, which can be calculated as:
\[ \binom{n}{x} = \frac{n!}{x!(n-x)!} \]

3. **Calculate the binomial coefficient:**
For our case:
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} \]
Simplifying this:
\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

4. **Calculate \(p^x\) and \(q^{n-x}\):**
- \(p^x = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\)
- \(q^{n-x} = \left(\frac{1}{2}\right)^{8-3} = \left(\frac{1}{2}\right)^5 = \frac{1}{32}\)

5. **Combine all parts to find the probability:**
\[ P(X = 3) = \binom{8}{3} \cdot p^3 \cdot q^{5} = 56 \cdot \frac{1}{8} \cdot \frac{1}{32} \]
Simplifying this:
\[ P(X = 3) = 56 \cdot \frac{1}{256} = \frac{56}{256} = \frac{7}{32} \]

6. **Final answer:**
The probability of getting exactly 3 heads when tossing 8 coins is \(\frac{7}{32}\).