Question

If a fair coin is tossed 8 times, then the probability that it shows heads more than tails is

• A. \( \frac{91}{256} \)
• B. \( \frac{97}{256} \)
• C. \( \frac{93}{256} \)
• D. \( \frac{95}{256} \)

Hint:

In problems involving binomial distributions, use the binomial probability formula to find the probability for different outcomes and sum them for the required range.

Answer 1

The Correct Option is C

Step 1: Use the binomial distribution formula.
When a fair coin is tossed \( n = 8 \) times, the probability of getting \( k \) heads is given by the binomial distribution formula: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] where \( p = \frac{1}{2} \), \( n = 8 \), and \( X \) is the number of heads.
Step 2: Calculate the number of outcomes for heads more than tails.
We need to find the probability that there are more heads than tails, which corresponds to \( X>4 \). This means we need to find the probability for \( X = 5, 6, 7, 8 \). Using the binomial distribution, we calculate: \[ P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \] This simplifies to \( \frac{93}{256} \).
Step 3: Conclusion.
Thus, the probability that heads appear more than tails is \( \frac{93}{256} \).