A fair coin is tossed 9 times. The probability that it shows heads in the first four tosses and tails in the last five tosses is

Options

A. `21/256`
B. `63/256`
C. `1/512`
D. `1/256`

Answer

To find the probability that a fair coin shows heads in the first four tosses and tails in the last five tosses when tossed 9 times, we can follow these steps:

### Step 1: Understand the problem
We need to calculate the probability of a specific sequence of outcomes when tossing a fair coin 9 times. The desired outcome is 4 heads followed by 5 tails.

### Step 2: Determine the probability for each toss
For a fair coin:
- The probability of getting heads (H) in one toss = \( \frac{1}{2} \)
- The probability of getting tails (T) in one toss = \( \frac{1}{2} \)

### Step 3: Calculate the probability for the first four tosses
Since we want heads in the first four tosses:
- The probability of getting heads in the first four tosses = \( \left(\frac{1}{2}\right)^4 \)

### Step 4: Calculate the probability for the last five tosses
Since we want tails in the last five tosses:
- The probability of getting tails in the last five tosses = \( \left(\frac{1}{2}\right)^5 \)

### Step 5: Combine the probabilities
Since the tosses are independent, we can multiply the probabilities of the two events:
\[ P(\text{4 heads and 5 tails}) = P(\text{4 heads}) \times P(\text{5 tails}) = \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right)^5 \]

### Step 6: Simplify the expression
\[ P(\text{4 heads and 5 tails}) = \left(\frac{1}{2}\right)^{4+5} = \left(\frac{1}{2}\right)^9 \]

### Step 7: Calculate the final probability
\[ P(\text{4 heads and 5 tails}) = \frac{1}{2^9} = \frac{1}{512} \]

Thus, the probability that the coin shows heads in the first four tosses and tails in the last five tosses is \( \frac{1}{512} \).

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