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Question

(A) Total number of ways to select two numbers from $100$ is ${}^{100}C_2 = \frac{100 	imes 99}{2} = 4950$.Let $S$ be the set $\{1, 2, 3, \dots, 100\

Vedclass · Accepted Answer

$0.55$

Vedclass · Answer

(A) Total number of ways to select two numbers from $100$ is ${}^{100}C_2 = \frac{100 	imes 99}{2} = 4950$.Let $S$ be the set $\{1, 2, 3, \dots, 100\}$. The numbers in $S$ divisible by $3$ are $\{3, 6, 9, \dots, 99\}$, which are $33$ in total. The numbers not divisible by $3$ are $100 - 33 = 67$.A product of two numbers is divisible by $3$ if at least one of the selected numbers is divisible by $3$.It is easier to calculate the complement: the product is $NOT$ divisible by $3$ if both selected numbers are $NOT$ divisible by $3$.The number of ways to select two numbers such that neither is divisible by $3$ is ${}^{67}C_2 = \frac{67 	imes 66}{2} = 2211$.The number of ways the product is divisible by $3$ is $4950 - 2211 = 2739$.The required probability is $\frac{2739}{4950} = 0.5533\dots$, which is $0.55$ correct to two decimal places.

Two numbers are selected at random from $1, 2, 3, \dots, 100$ and are multiplied. The probability, correct to two decimal places, that the product thus obtained is divisible by $3$ is:

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