Question

Three different numbers are selected at random from the set $A=\{1,2,3,......10\}$. The probability that the product of two of the numbers is equal to third is :

• A. $\frac{3}{4}$
• B. $\frac{1}{40}$
• C. $\frac{1}{8}$
• D. none of these

Answer 1

The correct option is B $\frac{1}{40}$

we have to select Three different numbers, such that the product of two of the numbers is equal to the third number

$2\times 3=6$

$2\times 4=8$

$2\times 5=10$

so 3 cases

so probility =$\frac{3}{{}^{10}{C}_3}$

$\frac{3}{120}=\frac{1}{40}$