Two distinct numbers are chosen from 1,3,5,7 ……. 151,153,155 and multiplied . The probability that the product is a multiple of 5 is

Options

A. `1020/3003`
B. `1112/3003`
C. `1011/3003`
D. `1122/3003`

Answer

To solve the problem, we need to find the probability that the product of two distinct numbers chosen from the sequence \(1, 3, 5, 7, \ldots, 155\) is a multiple of 5.

### Step-by-step Solution:

1. **Identify the Sequence**:
The sequence given is an arithmetic progression (AP) with the first term \(a = 1\) and a common difference \(d = 2\). The last term is \(155\).

2. **Find the Total Number of Terms**:
The general term of an AP is given by:
\[ T_n = a + (n-1)d \]
Setting \(T_n = 155\):
\[ 155 = 1 + (n-1) \cdot 2 \]
Simplifying this:
\[ 155 - 1 = (n-1) \cdot 2 \implies 154 = (n-1) \cdot 2 \implies n-1 = 77 \implies n = 78 \]
Thus, there are \(78\) terms in the sequence.

3. **Total Outcomes**:
The total ways to choose 2 distinct numbers from 78 is given by:
\[ \binom{78}{2} = \frac{78 \times 77}{2} = 3003 \]

4. **Identify Favorable Outcomes**:
We need to find the cases where the product of the two numbers is a multiple of 5. This occurs if at least one of the numbers is a multiple of 5.

5. **Find Multiples of 5 in the Sequence**:
The multiples of 5 in the sequence are \(5, 15, 25, 35, \ldots, 155\). This is also an AP with:
- First term \(5\)
- Common difference \(10\)
- Last term \(155\)

Setting the general term:
\[ T_n = 5 + (n-1) \cdot 10 \]
Setting \(T_n = 155\):
\[ 155 = 5 + (n-1) \cdot 10 \implies 150 = (n-1) \cdot 10 \implies n-1 = 15 \implies n = 16 \]
Thus, there are \(16\) multiples of 5 in the sequence.

6. **Calculate Favorable Outcomes**:
We can have two cases for favorable outcomes:
- **Case 1**: One number is a multiple of 5, and the other is not.
- Choose 1 from the 16 multiples of 5 and 1 from the 62 non-multiples of 5:
\[ \binom{16}{1} \cdot \binom{62}{1} = 16 \cdot 62 = 992 \]
- **Case 2**: Both numbers are multiples of 5.
- Choose 2 from the 16 multiples of 5:
\[ \binom{16}{2} = \frac{16 \cdot 15}{2} = 120 \]

Total favorable outcomes:
\[ 992 + 120 = 1112 \]

7. **Calculate the Probability**:
The probability \(P\) that the product is a multiple of 5 is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1112}{3003} \]

### Final Answer:
The probability that the product of two distinct numbers chosen from the sequence is a multiple of 5 is:
\[ \frac{1112}{3003} \]