How many ways are there to choose three distinct integers between 1 and 20 inclusive such that the numbers form an arithmetic sequence? *please try to answer by tomorrow/

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.