Question

Two numbers are selected at random from $1,2,3,.........100$ and multiplied. The probability that the product thus obtained is divisible by $3$ is ........

• A. $0.55$
• B. $0.62$
• C. $0.49$
• D. $0.60$

Answer 1

Answer: (A)

$0.55$

Total number of cases obtained by taking multiplication of anly two number 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 of there is multiplied with any one of remaining $67$ number or any two of these $33$ are multiplied, the resulting product is divisible by $3$.
Then the pair of number whose product is divisible by $3$ ${=}^{33}{C_1\times }^{67}{C}_1{+}^{33}{C}_2$
Hence the required probability $=\frac{{}^{33}{C}_1{\times }^{67}{C}_1{+}^{33}{C}_2}{{}^{100}{C}_2}=\frac{2739}{4950}=0.55$