What is the probability that a two-digit number selected at random will be a multiple of 3 and not a multiple of 5?

A- 2/15

B- 4/15

C- 1/15

D- 4/90

E- 8/11

What is the probability that a two-digit number selected at random will be a multiple of 3 and not a multiple of 5?

A- 2/15

B- 4/15

C- 1/15

D- 4/90

E- 8/11

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TopNewestThere are 30 2-digit numbers 12, 15, 18, , . . . . , 99 that are divisible by 3. Out of these 6 numbers, eg. 15, 30, 45. . . . , 90 are divisible by 5 as well. Hence remaining 24 numbers are favorable cases out of a total of 90 (from 10 to 99). Hence the required probability is 24/90 = 4/15 i.e.B..

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I think you missed one digit.

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Thanks Mehdi. I have corrected my mistake.

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