Agroup of six software packages available to solve a linear programming problem has been ranked from 1 to 6 (best to worst). an engineering firm, unaware of the rankings, randomly selected and then purchased two of the packages. let y denote the number of packages purchased by the firm that are ranked 3, 4, 5, or 6. give the probability distribution for y.

P(0) = 1/15 ≈ 0.067

P(1) = 8/15 ≈0.533

P(2) = 6/15 = 0.4

Step-by-step explanation:

Since only two of them were selected, there are only 3 possible numbers for Y: 0, 1 and 2.

0 indicates that non of the packages ranked 3, 4, 5, or 6 were chosen. There are only 2 possibilities were this is true: the first one is: first, package ranked 1 was purchased and then, package ranked 2 was purchased. The second possibility is: first, packaged ranked is purchased and then, package 2 is purchased. We can write this as (1,2) and (2,1). If we do not take onder into account this is going to be just (1,2), were we put arbitrarily the lower integer in the first place.

1 indicates that only one of the packages ranked 3, 4, 5, and 6 is chosen. In this case one of the purchased packages must be 1 or 2. So we have the 8 possibilites: (1,3) (1,4) (1,5) (1,6) (2,4) (2,5) (2,6).

2 indicates that 2 of the packages ranked 3, 4, 5 and 6 were chosen. So we are going to have 6 possibilites. (3,4) (3,5) (3,6) (4,5) (4,6) (5,6)

The total number of possibilities is 15.

The probability of Y being 0 is going to be P(0) = 1/15 ≈ 0.067

The probability of Y being 1 is going to be P(1) = 8/15 ≈0.533

The probability of Y being 2 is going to be P(2) = 6/15 = 0.4

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.