A bag contains 25 cards (numbered 1,2,3,4,………25) Two cards are picked at random from the bag (one after another and without replacement). What is the probability that the first card drawn has a prime number and the second card drawn has- a number which is multiple of 12?

Options

A. `3/100`
B. `1/50`
C. `1/25`
D. `2/75`

Answer

To solve the problem step by step, we will calculate the probability that the first card drawn has a prime number and the second card drawn has a number which is a multiple of 12.

### Step 1: Identify the prime numbers between 1 and 25
The prime numbers between 1 and 25 are:
- 2, 3, 5, 7, 11, 13, 17, 19, 23

**Total prime numbers = 9**

### Step 2: Identify the multiples of 12 between 1 and 25
The multiples of 12 between 1 and 25 are:
- 12, 24

**Total multiples of 12 = 2**

### Step 3: Calculate the probability of drawing a prime number first
The probability of drawing a prime number first can be calculated as:
\[ P(\text{Prime}) = \frac{\text{Number of prime numbers}}{\text{Total cards}} = \frac{9}{25} \]

### Step 4: Calculate the probability of drawing a multiple of 12 second
Since we are drawing without replacement, after drawing the first card (which is a prime number), there will be 24 cards left in the bag. The favorable outcomes for the second draw (drawing a multiple of 12) remain the same since neither 12 nor 24 is a prime number.

Thus, the probability of drawing a multiple of 12 second is:
\[ P(\text{Multiple of 12 | Prime drawn}) = \frac{\text{Number of multiples of 12}}{\text{Total remaining cards}} = \frac{2}{24} = \frac{1}{12} \]

### Step 5: Calculate the combined probability
Since we want the probability that both events occur (drawing a prime number first and a multiple of 12 second), we multiply the probabilities of the two independent events:
\[ P(\text{Prime and Multiple of 12}) = P(\text{Prime}) \times P(\text{Multiple of 12 | Prime drawn}) = \frac{9}{25} \times \frac{1}{12} \]

### Step 6: Simplify the combined probability
Now we simplify:
\[ P(\text{Prime and Multiple of 12}) = \frac{9 \times 1}{25 \times 12} = \frac{9}{300} \]
This can be reduced:
\[ \frac{9}{300} = \frac{3}{100} \]

### Final Answer
The required probability that the first card drawn has a prime number and the second card drawn has a number which is a multiple of 12 is:
\[ \frac{3}{100} \]