Question

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

• A. $\frac{1020}{3003}$
• B. $\frac{1112}{3003}$
• C. $\frac{1011}{3003}$
• D. $\frac{1122}{3003}$

Answer 1

Answer: (B)

Given numbers $1,3,\mathrm{5........},153,155$
$T_n=a+\left(n-1\right)d\Rightarrow 155=1+\left(n-1\right)2\Rightarrow n=78$
Total number of ways ${=}^{78}{C}_2=3003$
The numbers which are multiple of $5$ are $5,15,25,....,155$
$155=5+\left(n-1\right)10\Rightarrow n=16$
The number of ways in which product is a multiple of $5=$ (Both two numbers from $5,15,25,......,155$ ) or (one number from $5,15,25,......,145$ and one number from the remaining numbers).
${=}^{16}{C}_2{+}^{16}{C}_1{\times }^{78-16}{C}_1$
$=\frac{816\times 15}{2}+16\times 62$
$=120+992=1112$