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Paul and Jesse each choose a number at random from the first six primes. What is the probability that the sum of the numbers they choose is a multiple of 3?

 Nov 19, 2020
 #1
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Primes[2, 3, 5, 7, 11, 13]

 

(5, 7, 8, 9, 10, 12, 13, 14, 15, 16, 16, 18, 18, 20, 24) >>Total Sums=  15

 

[9, 12, 15, 18, 18, 24]>>Total = 6 - numbers that are divisible by 3.

 

Probability is: 6 / 15 =2 / 5

 Nov 19, 2020
 #2
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I think this is more accurate:

 

(4, 5, 6, 7, 8, 9, 10, 10, 12, 13, 14, 14, 15, 16, 16, 18, 18, 20, 22, 24, 26)>>Total Sums =21

 

[6, 9, 12, 15, 18, 18, 24] = 7 Numbers divisible by 3

 

Probability is: 7 / 21 = 1 / 3

Guest Nov 19, 2020

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