A red box contains eight items , of which three are defective, and a blue box contains five items, of which two are defective . An item is drawn at random from each box. What is the probability that one item is defective and one is not ?

Options

A. `(17)/(20)`
B. `(5)/(8)`
C. `(17)/(32)`
D. `(19)/(40)`

Explanation

To solve the problem, we need to calculate the probability that one item drawn from the red box is defective and the item drawn from the blue box is not defective, or vice versa.

### Step-by-Step Solution:

1. **Identify the total items and defective items in each box:**
- Red box: 8 items (3 defective, 5 non-defective)
- Blue box: 5 items (2 defective, 3 non-defective)

2. **Calculate the probabilities:**
- Probability of drawing a defective item from the red box:
\[ P(\text{Defective from Red}) = \frac{3}{8} \]
- Probability of drawing a non-defective item from the red box:
\[ P(\text{Non-defective from Red}) = \frac{5}{8} \]
- Probability of drawing a defective item from the blue box:
\[ P(\text{Defective from Blue}) = \frac{2}{5} \]
- Probability of drawing a non-defective item from the blue box:
\[ P(\text{Non-defective from Blue}) = \frac{3}{5} \]

3. **Calculate the probability of one defective and one non-defective item:**
- There are two scenarios to consider:
- Scenario 1: Defective from Red and Non-defective from Blue
- Scenario 2: Non-defective from Red and Defective from Blue

- For Scenario 1:
\[ P(\text{Defective from Red and Non-defective from Blue}) = P(\text{Defective from Red}) \times P(\text{Non-defective from Blue}) = \frac{3}{8} \times \frac{3}{5} = \frac{9}{40} \]

- For Scenario 2:
\[ P(\text{Non-defective from Red and Defective from Blue}) = P(\text{Non-defective from Red}) \times P(\text{Defective from Blue}) = \frac{5}{8} \times \frac{2}{5} = \frac{10}{40} \]

4. **Add the probabilities of both scenarios:**
\[ P(\text{One defective and one non-defective}) = P(\text{Defective from Red and Non-defective from Blue}) + P(\text{Non-defective from Red and Defective from Blue} \]
\[ = \frac{9}{40} + \frac{10}{40} = \frac{19}{40} \]

5. **Final Answer:**
The probability that one item is defective and one is not is:
\[ \boxed{\frac{19}{40}} \]