In a ward-robe, Nitish has 3 trousers. One of them is black, second is blue and third brown. In this wardrobe, he has 4 shirts also. One of them is black and the other 3 are white. He opens his ward-robe in the dark and picks out one shirt-trouser pair without examining the colour. What is the likelihood that neither the shirts nor the trousers are black?

Options

A. a. `1/12`
B. b. `1/6`
C. c. `1/4`
D. d. `1/2`

Answer

To solve the problem, we need to find the probability that neither the shirt nor the trouser picked by Nitish is black. Let's break it down step by step.

### Step 1: Identify the total number of trousers and shirts
- **Trousers**: Nitish has 3 trousers:
- 1 Black
- 1 Blue
- 1 Brown
- **Shirts**: Nitish has 4 shirts:
- 1 Black
- 3 White

### Step 2: Calculate the total combinations of shirt-trouser pairs
The total number of combinations of trousers and shirts can be calculated by multiplying the number of trousers by the number of shirts:
\[ \text{Total combinations} = \text{Number of trousers} \times \text{Number of shirts} = 3 \times 4 = 12 \]

### Step 3: Calculate the number of favorable outcomes
We need to find the combinations where neither the shirt nor the trouser is black.

- **Non-black trousers**: There are 2 non-black trousers (Blue and Brown).
- **Non-black shirts**: There are 3 non-black shirts (3 White).

Now, we can calculate the number of favorable combinations:
\[ \text{Favorable combinations} = \text{Non-black trousers} \times \text{Non-black shirts} = 2 \times 3 = 6 \]

### Step 4: Calculate the probability
The probability that neither the shirt nor the trouser is black can be calculated using the formula:
\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{6}{12} = \frac{1}{2} \]

### Final Answer
Thus, the probability that neither the shirt nor the trouser is black is:
\[ \frac{1}{2} \]

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