A fair coin is tossed four times. What is the probability that at most three tails occur?

Options

A. `(7)/(8)`
B. `(15)/(16)`
C. `(13)/(16)`
D. `(3)/(4)`

Answer

To solve the problem of finding the probability that at most three tails occur when a fair coin is tossed four times, we can follow these steps:

### Step 1: Determine the total number of outcomes
When a coin is tossed, there are two possible outcomes: heads (H) or tails (T). If the coin is tossed \( n \) times, the total number of outcomes can be calculated using the formula \( 2^n \).

For our case:
- \( n = 4 \) (the coin is tossed four times)
- Total outcomes = \( 2^4 = 16 \)

### Step 2: Identify the favorable outcomes
We need to find the number of outcomes where at most three tails occur. "At most three tails" means we can have 0, 1, 2, or 3 tails.

1. **0 tails (all heads)**: There is only 1 outcome: HHHH
2. **1 tail**: The number of ways to choose 1 tail from 4 tosses is given by \( \binom{4}{1} \):
\[ \binom{4}{1} = 4 \]
(HTHH, HHTH, HTHH, HHHT)

3. **2 tails**: The number of ways to choose 2 tails from 4 tosses is given by \( \binom{4}{2} \):
\[ \binom{4}{2} = 6 \]
(TTHH, THTH, THHT, HTTH, HHTT, HHTH)

4. **3 tails**: The number of ways to choose 3 tails from 4 tosses is given by \( \binom{4}{3} \):
\[ \binom{4}{3} = 4 \]
(TTTH, TTHT, THTT, HTTT)

Now, we can sum these outcomes:
- Outcomes with 0 tails: 1
- Outcomes with 1 tail: 4
- Outcomes with 2 tails: 6
- Outcomes with 3 tails: 4

Total favorable outcomes = \( 1 + 4 + 6 + 4 = 15 \)

### Step 3: Calculate the probability
The probability of an event is given by the formula:
\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \]

Substituting the values we found:
\[ \text{Probability} = \frac{15}{16} \]

### Final Answer
The probability that at most three tails occur when a fair coin is tossed four times is \( \frac{15}{16} \).

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