If two numbers are chosen at random from 1, 2 | vedclass

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(A) The total number of ways to choose $2$ numbers from $100$ is $^{100}C_2 = \frac{100 	imes 99}{2} = 4950$.Among the numbers from $1$ to $100$,the

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$0.55$

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(A) The total number of ways to choose $2$ numbers from $100$ is $^{100}C_2 = \frac{100 	imes 99}{2} = 4950$.Among the numbers from $1$ to $100$,the numbers divisible by $3$ are $3, 6, 9, \dots, 99$. There are $33$ such numbers.The numbers not divisible by $3$ are $100 - 33 = 67$.The product of two numbers is divisible by $3$ if at least one of the chosen numbers is a multiple of $3$.This can happen in two cases:Case $1$: One number is a multiple of $3$ and the other is not. Number of ways = $^{33}C_1 	imes ^{67}C_1 = 33 	imes 67 = 2211$.Case $2$: Both numbers are multiples of $3$. Number of ways = $^{33}C_2 = \frac{33 	imes 32}{2} = 528$.Total favorable outcomes = $2211 + 528 = 2739$.Probability = $\frac{2739}{4950} = 0.55$.

If two numbers are chosen at random from $1, 2, 3, \dots, 100$ and multiplied,what is the probability that the product is divisible by $3$?

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