The probability that A hits a target is `1/3` and the probability that B hits It is `2/5` . What is the probability that the target will be hit , if each one of A and B shoots the target ?

Options

A. `5/6`
B. `3/5`
C. `11/15`
D. `1/6`

Answer

To solve the problem, we need to find the probability that at least one of A or B hits the target when both shoot at it. We can use the formula for the union of two probabilities.

### Step-by-Step Solution:

1. **Identify the probabilities:**
- The probability that A hits the target, \( P(A) = \frac{1}{3} \)
- The probability that B hits the target, \( P(B) = \frac{2}{5} \)

2. **Use the formula for the probability of the union of two events:**
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Here, \( P(A \cup B) \) is the probability that at least one of them hits the target.

3. **Calculate \( P(A \cap B) \):**
- Since A and B are independent events, we can find the probability that both A and B hit the target by multiplying their probabilities:
\[ P(A \cap B) = P(A) \times P(B) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15} \]

4. **Substitute the values into the union formula:**
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{1}{3} + \frac{2}{5} - \frac{2}{15} \]

5. **Find a common denominator:**
- The least common multiple (LCM) of 3, 5, and 15 is 15. We will convert each fraction:
\[ P(A) = \frac{1}{3} = \frac{5}{15}, \quad P(B) = \frac{2}{5} = \frac{6}{15}, \quad P(A \cap B) = \frac{2}{15} \]

6. **Perform the addition and subtraction:**
\[ P(A \cup B) = \frac{5}{15} + \frac{6}{15} - \frac{2}{15} = \frac{5 + 6 - 2}{15} = \frac{9}{15} \]

7. **Simplify the result:**
\[ P(A \cup B) = \frac{9}{15} = \frac{3}{5} \]

### Final Answer:
The probability that the target will be hit if each one of A and B shoots the target is \( \frac{3}{5} \).

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