Question

Two numbers are selected at random from $1,2,3,....,100$ and are multiplied, then the probability (correct to two places of decimals) that the product thus obtained is divisible by $3$ is

• A. $0.22$
• B. $0.33$
• C. $0.44$
• D. $0.55$

Answer 1

The correct option is D $0.55$

Total number of cases obtained by taking multiplication of only two numbers out of $100={}^{100}{C}_2$
Out of $100(1,2,...,100)$ given numbers, there are the numbers $3,6,9,\mathrm{12...},99$
which are $33$ in numbers such that when any one of them is multiplied with any of the remaining $67$ numbers
as any two of these $33$ are multiplied, the resulting product is divisible by $3$.
Then the pairs of numbers whose product is divisible by $3{=}^{33}{C}_1{\times }^{67}{C}_1{+}^{33}{C}_2$
Hence, the required probability$=\frac{{}^{33}{C}_1{\times }^{67}{C}_1{+}^{33}{C}_2}{{}^{100}{C}_2}=\frac{2739}{4950}=0.55$