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Question

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

Vedclass · Accepted Answer

$0.55$

Vedclass · Answer

(A) Total number of ways to select two distinct numbers from $100$ is ${}^{100}C_2 = \frac{100 	imes 99}{2} = 4950$.Let $S$ be the set ${1, 2, \dots, 100}$. The numbers in $S$ divisible by $3$ are ${3, 6, \dots, 99}$,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 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 $0.55$ when rounded to two decimal places.

$A$. Probability that none of the balls is white	$I$. $\frac{1}{70}$
$B$. Probability of getting $2$ white and $1$ blue ball	$II$. $\frac{6}{35}$
$C$. Probability of getting $2$ blue and $1$ white ball	$III$. $\frac{13}{20}$
$D$. Probability of getting $1$ red, $1$ white and $1$ blue ball	$IV$. $\frac{1}{10}$

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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