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What is the probability that a number selected at random from the set {2, 3, 7, 12, 15, 22, 72, 108} will be divisible by both 2 and 3 ?
F. 1/4
G. 3/8
H. 3/5
J. 5/8
K. 7/8


Sagot :

Total amount of numbers in the set = 8 .

Total amount that are divisible by both 2 and 3 = 3 .

So the probability of picking one of them is 3/8 = 37.5 percent.
[tex]|\Omega|=8\\ |A|=3\\ P(A)=\frac{|A|}{|\Omega|}\\ P(A)=\frac{3}{8}[/tex]