Question

The probability of getting 4 heads in 6 tosses of a fair coin is.

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{64} \)
• C. \( \frac{15}{64} \)
• D. \( \frac{3}{64} \)

Hint:

For a binomial distribution, use the formula \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \), where \( \binom{n}{k} \) represents the number of ways to choose \( k \) successes from \( n \) trials.

Answer 1

The Correct Option is C

To calculate the probability of getting 4 heads in 6 tosses of a fair coin, we use the binomial distribution formula:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] Where:
- \( n = 6 \) is the number of tosses,
- \( k = 4 \) is the number of heads,
- \( p = \frac{1}{2} \) is the probability of getting heads in one toss,
- \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) heads from \( n \) tosses.
Now, calculate the binomial coefficient:
\[ \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \] So, the probability is:
\[ P(X = 4) = \binom{6}{4} \left( \frac{1}{2} \right)^4 \left( \frac{1}{2} \right)^{6-4} \] \[ P(X = 4) = 15 \times \left( \frac{1}{2} \right)^6 = 15 \times \frac{1}{64} = \frac{15}{64} \]
Conclusion: The probability of getting 4 heads in 6 tosses of a fair coin is \( \frac{15}{64} \).