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

Options

A. `31/128`
B. `93/128`
C. `31/256`
D. `93/256`

Answer

To solve the problem of finding the probability that a fair coin tossed 8 times shows heads more often than tails, we can follow these steps:

### Step 1: Understand the problem
We need to determine the probability that the number of heads (H) is greater than the number of tails (T) when a fair coin is tossed 8 times. This means we are looking for the cases where the number of heads is greater than 4.

### Step 2: Define the random variable
Let \( X \) be the random variable representing the number of heads obtained in 8 tosses. Since the coin is fair, \( X \) follows a binomial distribution with parameters \( n = 8 \) (number of trials) and \( p = 0.5 \) (probability of getting heads in each trial).

### Step 3: Determine the required probability
We need to find \( P(X > 4) \). This can be calculated as:
\[ P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \]

### Step 4: Calculate each probability using the binomial formula
The probability mass function of a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
For our case, \( n = 8 \) and \( p = 0.5 \).

1. **Calculate \( P(X = 5) \)**:
\[ P(X = 5) = \binom{8}{5} \left(0.5\right)^5 \left(0.5\right)^{3} = \binom{8}{5} \left(0.5\right)^8 \]
\[ = \frac{8!}{5!3!} \left(0.5\right)^8 = \frac{56}{256} = \frac{7}{32} \]

2. **Calculate \( P(X = 6) \)**:
\[ P(X = 6) = \binom{8}{6} \left(0.5\right)^6 \left(0.5\right)^{2} = \binom{8}{6} \left(0.5\right)^8 \]
\[ = \frac{8!}{6!2!} \left(0.5\right)^8 = \frac{28}{256} = \frac{7}{64} \]

3. **Calculate \( P(X = 7) \)**:
\[ P(X = 7) = \binom{8}{7} \left(0.5\right)^7 \left(0.5\right)^{1} = \binom{8}{7} \left(0.5\right)^8 \]
\[ = \frac{8!}{7!1!} \left(0.5\right)^8 = \frac{8}{256} = \frac{1}{32} \]

4. **Calculate \( P(X = 8) \)**:
\[ P(X = 8) = \binom{8}{8} \left(0.5\right)^8 \left(0.5\right)^{0} = \binom{8}{8} \left(0.5\right)^8 \]
\[ = 1 \cdot \left(0.5\right)^8 = \frac{1}{256} \]

### Step 5: Sum the probabilities
Now we sum the probabilities calculated:
\[ P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \]
\[ = \frac{7}{32} + \frac{7}{64} + \frac{1}{32} + \frac{1}{256} \]
To add these fractions, we convert them to a common denominator (256):
\[ = \frac{56}{256} + \frac{28}{256} + \frac{8}{256} + \frac{1}{256} \]
\[ = \frac{56 + 28 + 8 + 1}{256} = \frac{93}{256} \]

### Final Answer
Thus, the probability that the coin shows heads more often than tails when tossed 8 times is:
\[ \boxed{\frac{93}{256}} \]