Answer:
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
It is a mixed number because 12/2 can be written as 6 instead of 12/2
Answer:
Infinite number of solutions.
Step-by-step explanation:
Are the equations written correctly? They are identical. Since they overlap, every point is a solution for as long as the lines reach. Since no one is stopping them, the answers are infinite.
