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Question

Two numbers are selected at random from 1,2,3,....,100 and are multiplied, then the probability (correct to two places of decimals) that the product thus obtained is divisible by 3 is

A
0.22
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B
0.33
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C
0.44
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D
0.55
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Solution

The correct option is D 0.55
Total number of cases obtained by taking multiplication of only two numbers out of 100=100C2
Out of 100(1,2,...,100) given numbers, there are the numbers 3,6,9,12...,99
which are 33 in numbers such that when any one of them is multiplied with any of the remaining 67 numbers
as any two of these 33 are multiplied, the resulting product is divisible by 3.
Then the pairs of numbers whose product is divisible by 3=33C1×67C1+33C2
Hence, the required probability=33C1×67C1+33C2100C2=27394950=0.55

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