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Question

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

A
13
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B
2366
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C
12
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D
none of these
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Solution

## The correct option is B 13First12naturalnumbersare1,2,3,........,12Numberofwaysinwhichtwodistinctnumberscanbeselectedoutofthese12naturalnumbers=(122)LetP(T)=(122)Themaximumsumoftwoselecteddistinctnumbers=12+11=23Theminimumsumoftwoselecteddistinctnumbers=1+2=3Forthesumbeingdivisibleby3itmustbeoneof3,6,9,12,15,18,21LetSbetheeventofsumoftwoselectednumber.P(S=3)=1P(T)P(S=6)=2P(T)P(S=9)=4P(T)P(S=12)=5P(T)P(S=15)=5P(T)P(S=18)=3P(T)P(S=21)=2P(T)LetDbetheeventthatsumisdivisibleby3ThenP(D)=P(S=3)+P(S=6)+P(S=9)+P(S=12)+P(S=15)+P(S=18)+P(S=21)⇒P(D)=22P(T)=2266⇒P(D)=13

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