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29 Feb 2016, 09:40
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If a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20, what is the probability that this product will be a multiple of 45?
A. 1/32
B. 1/28
C. 1/24
D. 1/16
E. 1/14
A. 1/32
B. 1/28
C. 1/24
D. 1/16
E. 1/14
FightToSurvive
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29 Feb 2016, 10:22
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If a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20, what is the probability that this product will be a multiple of 45?
A. 1/32
B. 1/28
C. 1/24
D. 1/16
E. 1/14
There are 3 single digit multiple of 3, that is, 3,6,9.
There are 8 prime nos less than 20 - 2,3,5,7,11,13,17,19
Total outcome - 8*3 = 24
Favourable outcome = 1 (9*5)
Hence required probability = 1/24. Answer C.
A. 1/32
B. 1/28
C. 1/24
D. 1/16
E. 1/14
There are 3 single digit multiple of 3, that is, 3,6,9.
There are 8 prime nos less than 20 - 2,3,5,7,11,13,17,19
Total outcome - 8*3 = 24
Favourable outcome = 1 (9*5)
Hence required probability = 1/24. Answer C.
12 Sep 2019, 14:06
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Bunuel
There are 4 single digit multiples of 3 --> 0,3,6,9 [Since, 0 is a multiple of 3]
There are 8 prime numbers less than 20 --> 2,3,5,7,11,13,17,19
Total number of multiplications = 8*4 = 32
Favorable outcomes = (All multiplications involving 0) AND (9*5) = 8+1 = 9 [Since, 0 is a multiple of 45]
Hence, the required probability is 9/32
Bunuel What am I getting wrong here?
