VeritasPrepKarishma wrote:

CCMBA wrote:

I did the lengthier solution, but I get something different.

1-[Pr(Neither) + Pr(J, not P) + Pr(P, not J)]

1 - [3/18+ 2/18]

13/18

What did I do wrong?

Ex: Pr(Neither) = 7/9*6/8...*3/5 = 1/6

P(Both not selected) = (4/9) * (3/8) = 1/6 (The probability of John being among the 4 leftover out of 9 people is 4/9 and probability of Peter being among the leftover 3 is 3/8)

P(Peter selected, John is not) = (5/9)*(4/8) = 5/18 (Probability of Peter being among 5 selected is 5/9 and John being among the leftover 4 is 4/8)

P(Peter not selected, John is selected) = (5/9)*(4/8) = 5/18 (Probability of John being among 5 selected is 5/9 and Peter being among the leftover 4 is 4/8)

1 - (1/6 + 5/18 + 5/18) = 5/18

Hi Karishma,

Thank you for your response. There are still 2 things I do not understand.

The first is the fractions you use. Are we taking as given that the first 3 spots are filled? So for Pr(Neither), we're only considering the last spot, and we want to ignore J and P?

Second, where is the flaw in the method I used?

Pr(Neither) = 7/9 * 6/8 * 5/7 * 4/6 * 3/5 = 1/6

Pr(J, not P) = Pr(P, not J) = 1/9 * 7/8 * 6/7 * 5/6 * 4/5 = 1/18

Pr(J) is 1/9. Then we are only considering 7 of the remaining 8 people, then 6 of 7, etc.

Thank you.