A soldier is firing at a moving target. He fires four shots. The probability of hitting the target at the first, second, third and fourth shots are 0.6, 0.4, 0.2 and 0.1 respectively. What is the probability that he hits the target?

Options

A. `527/625`
B. `3/625`
C. `105/625`
D. none of these

Answer

To find the probability that the soldier hits the target at least once after firing four shots, we can use the concept of complementary probability. We'll calculate the probability that he misses all four shots and then subtract that from 1.

### Step-by-Step Solution:

1. **Identify the probabilities of hitting the target for each shot:**
- Probability of hitting on the first shot, \( P(A) = 0.6 \)
- Probability of hitting on the second shot, \( P(B) = 0.4 \)
- Probability of hitting on the third shot, \( P(C) = 0.2 \)
- Probability of hitting on the fourth shot, \( P(D) = 0.1 \)

2. **Calculate the probabilities of missing the target for each shot:**
- Probability of missing the first shot, \( P(A') = 1 - P(A) = 1 - 0.6 = 0.4 \)
- Probability of missing the second shot, \( P(B') = 1 - P(B) = 1 - 0.4 = 0.6 \)
- Probability of missing the third shot, \( P(C') = 1 - P(C) = 1 - 0.2 = 0.8 \)
- Probability of missing the fourth shot, \( P(D') = 1 - P(D) = 1 - 0.1 = 0.9 \)

3. **Calculate the probability of missing all four shots:**
- The probability of missing all shots is the product of the probabilities of missing each shot:
\[ P(A' \cap B' \cap C' \cap D') = P(A') \times P(B') \times P(C') \times P(D') \]
\[ = 0.4 \times 0.6 \times 0.8 \times 0.9 \]

4. **Perform the multiplication:**
- First, calculate \( 0.4 \times 0.6 = 0.24 \)
- Next, calculate \( 0.24 \times 0.8 = 0.192 \)
- Finally, calculate \( 0.192 \times 0.9 = 0.1728 \)

5. **Calculate the probability of hitting the target at least once:**
- The probability of hitting the target at least once is:
\[ P(\text{hit at least once}) = 1 - P(A' \cap B' \cap C' \cap D') \]
\[ = 1 - 0.1728 = 0.8272 \]

6. **Express the final answer as a fraction:**
- To express \( 0.8272 \) as a fraction, we can convert it:
\[ 0.8272 = \frac{8272}{10000} = \frac{517}{625} \]

### Final Answer:
The probability that the soldier hits the target at least once is \( \frac{517}{625} \).