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

1

step-by-step explanation:

it's the only multiple of both and the least

ruth is 6 years old

step-by-step explanation:

6 + 24 = 30,   so 6/30 = .2  

if you do 1/5 it also equals .2, so that is the answer.