Answer:
probability of the product of the chosen integers being a multiple of 3 is P(E)= 1 - (91/285)
=194/285 or 0.6807.
Step-by-step explanation:
The sample space of all possible choices of three integers from the set {1,2,…..,20} has C(20,3) = (20×19×18)/3! elements.
The complement of the event space E consists of all possible choices of three integers from the complement of the set of all the multiples of 3 in the above set because the product of the chosen integers is a multiple of 3 if and only if at least one of them is a multiple of 3. Hence we have to choose three elements from the set {1,2,4,5,7,8,10,11,13,14,16,17,19,20} which has 14 elements.
Hence
|E'| = C(14,3)
= 14×13×12/3!.
Therefore probability P(E')
= |E'|/|S|
= (14×13×12)/(20×19×18)
= (14×13×2)/(20×19×3)
=(7×13)/(5×19×3)
= 91/285.
Therefore the required probability of the product of the chosen integers being a multiple of 3 is P(E)= 1 - (91/285)=194/285 or 0.6807.
