Question

A coin is tossed 10 times. The probability of getting exactly six heads is:

• A. \( \binom{10}{6} \left( \frac{1}{2} \right)^6 \left( \frac{1}{2} \right)^4 \)
• B. \( \binom{10}{6} \left( \frac{1}{2} \right)^7 \)
• C. \( \binom{10}{6} \left( \frac{1}{2} \right)^8 \)
• D. \( \binom{10}{6} \left( \frac{1}{2} \right)^{10} \)

Hint:

Use the binomial formula for problems involving a fixed number of successes in repeated independent trials: \[ P(r) = \binom{n}{r} p^r q^{n-r} \] where \( p \) is the probability of success, and \( q = 1 - p \).

Answer 1

The Correct Option is D

Step 1: This is a binomial probability problem. The probability of getting exactly \( r \) heads in \( n \) tosses is given by: \[ P = \binom{n}{r} p^r q^{n-r} \] Step 2: For a fair coin: - \( n = 10 \) - \( r = 6 \) - \( p = q = \frac{1}{2} \) Step 3: Substitute values: \[ P = \binom{10}{6} \left( \frac{1}{2} \right)^6 \left( \frac{1}{2} \right)^4 = \binom{10}{6} \left( \frac{1}{2} \right)^{10} \] Final Answer: \( \boxed{\binom{10}{6} \left( \frac{1}{2} \right)^{10}} \)