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  • Two distinct numbers are selected at random from the first twelve natural numbers. The probability that the sum will be divisible by 3 isOption: 1 <i

Two distinct numbers are selected at random from the first twelve natural numbers. The probability that the sum will be divisible by 3 is

Option: 1

\frac{1}{3}


Option: 2

\frac{23}{66}


Option: 3

\frac{1}{2}


Option: 4

None of these


Answers (1)

best_answer

Let \mathrm{\mathrm{E}_3=} the event of the sum being 3 . Similarly, \mathrm{\mathrm{E}_6, \mathrm{E}_9, \mathrm{E}_{12}, \mathrm{E}_{15}, \mathrm{E}_{18}, \mathrm{E}_{21}.}

\begin{aligned} & \mathrm{n(S)={ }^{12} C_2 }\\ \\& \mathrm{n\left(E_3\right)=1, n\left(E_6\right)=2, n\left(E_9\right)=4, n\left(E_{12}\right)=5, n\left(E_{15}\right)=5, n\left(E_{18}\right)=3, n\left(E_{21}\right)=2 .} \\ \\& \therefore \quad \mathrm{P(E)=P\left(E_3\right)+\ldots+P\left(E_{21}\right)=\frac{1+2+4+5+5+3+2}{{ }^{12} C_2}=\frac{22 \times 2}{12 \times 11}=\frac{1}{3} }. \end{aligned}

Alternatively The sum of the numbers for every selection is divisible by 3 or leaves the remainder 1 or leaves the remainder 2 . These are equally probable. So, the required probability =\frac{1}{3}, because the sum of the three probabilities is 1 .

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avinash.dongre

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