g A law firm has six senior and seven junior partners. A committee of three partners is selected at random to represent the firm at a conference. What is the probability that at least one of the junior partners is on the committee?

133/143

Step-by-step explanation:

Let S be the sample space

Let E be the event of selecting three committee partners with at least one junior partner.

Partners in the law firm include:

Senior partners = 6

Junior partners = 7

Total partners = 13

n (S) = number of ways of selecting 3 partners from 13 = 13C3

n (S) = 13C3 = 13! / (10!3!) = (13x12x11) / (3x2x1) = 286

To get n (E) i. e least 1 junior partner in the selected committee, we may have:

(2 senior and 1 junior) or (1 senior and 2 junior) or (3 junior).

Therefore, the required number of way is given below:

= (6C2 x 7C1) + (6C1 x 7C2) + 7C3

= [ (6x5) / 2 x 7] + [6 x (7x6) / 2] + [ (7x6x5) / (3x2) ]

= 105 + 126 + 35

n (E) = 266

Therefore, the probability P (E) that at least one of the junior partners is on the committee is given below:

P (E) = n (E) / n (S)

P (E) = 266/286

P (E) = 133/143