Question

A coin is thrown 8 times. What is the probability of getting a head in an odd number of throws?

• A. \( \frac{3}{4} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{8} \)

Hint:

For binomial distributions, the probability of getting an odd number of successes is often symmetric when \( p = 0.5 \).

Answer 1

The Correct Option is C

Step 1: Define the problem. The coin is thrown 8 times, so the total number of outcomes is \( 2^8 = 256 \). We are asked to find the probability of getting heads in an odd number of throws. 
Step 2: Use the binomial distribution. The number of heads in 8 throws follows a binomial distribution with parameters \( n = 8 \) and \( p = 0.5 \). The probability of getting an odd number of heads is the sum of the probabilities of getting 1, 3, 5, or 7 heads. The probability of getting \( r \) heads is given by the binomial probability: \[ P(r) = \binom{8}{r} \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8-r} = \binom{8}{r} \left( \frac{1}{2} \right)^8. \] Step 3: Calculate the probabilities. The probability of getting an odd number of heads is the sum of probabilities for \( r = 1, 3, 5, 7 \): \[ P({odd heads}) = P(1) + P(3) + P(5) + P(7). \] Using binomial coefficients: \[ P({odd heads}) = \frac{1}{256} \left( \binom{8}{1} + \binom{8}{3} + \binom{8}{5} + \binom{8}{7} \right). \] The sum of these binomial coefficients is 128, so the probability is: \[ P({odd heads}) = \frac{128}{256} = \frac{1}{2}. \] Thus, the correct answer is: \[ \boxed{\frac{1}{2}}. \]