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Difficult Math Question #33 - Probability

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by 800guy » Mon Oct 23, 2006 2:27 pm
A number is selected at random from first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 13?

(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30
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by rajs.kumar » Tue Oct 24, 2006 2:49 am
2/5.

10 multiples of 3 and 2 multiples of 13 adding to a total of 12. So the probability is 12/30.
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by ajith » Tue Oct 24, 2006 8:34 pm
rajs.kumar wrote:2/5.

10 multiples of 3 and 2 multiples of 13 adding to a total of 12. So the probability is 12/30.
Agree there
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by limits660 » Wed Oct 25, 2006 4:23 am
I believe its B
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OA

by 800guy » Wed Oct 25, 2006 3:29 pm
here's the OA:
Total no from 1 to 30 = 30
total no from 1 to 30 which r multiple of 3 = 10 (eg(3,6,9,12,15,18,21,24,27,30))
total no from 1 to 30 which r multiple of 13 = 2 (eg 13,26)
P(a or b ) = p(a) + p(b)
p(a)= 10/30
p(b)=2/30
p(a) + p(b) = 10/30+2/30 = 2/5
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