Alaw firm has five senior and six 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?

the probability that at least one of the junior partners is on the committee is 0.9393

Step-by-step explanation:

The probability that a event happens is calculate as the quotient between the numbers of possibilities in which that event happens and the numbers of total possibilities.

Then for calculate the number of possibilities is necessary to introduce the concept of combination. The number of combinations are the ways that we can form groups with size k from a bigger group of size n and it is calculate as:

nCk=n!/(k!(n-k)! )

In other words nCk give as the number of ways that we can form groups of k objects from a group of n objects.

Taking into account the last explanation, the number of groups of 3 that we can create with the 11 partners (5 junior and 6 senior) is 165 and is calculate as:

11C3= 11!/(3!(11-3)! )

11C3=165

Then, for have at least 1 junior partner, we have 3 options:

1.A group formed by 1 junior partner and 2 senior partner

2.A group formed by 2 junior partner and 1 senior partner

3.A group formed by 3 junior partner and 0 senior partner

For the first option: we have to find the number of ways to choose one junior partner from 6 general options and multiply to the number of ways to choose two senior partner from 5 general options. This is calculate as:

Possibilities 1 = 6C1*5C2 = 6*10 = 60

That mean that there are 60 ways to choose group of 3 formed by one junior partner and 2 senior partners.

For the second option: we have to find the number of ways to choose two junior partner from 6 general options and multiply to the number of ways to choose one senior partner from 5 general options. This is calculate as:

Possibilities 2 = 6C2*5C1 = 15*5 = 75

That mean that there are 75 ways to choose group of 3 formed by 2 junior partner and 1 senior partners.

For the third option: we have to find the number of ways to choose three junior partner from 6 general options and multiply to the number of ways to choose zero partner from 5 general options. This is calculate as:

Possibilities 3 = 6C3*5C0 = 20*1 = 20

That mean that there are 20 ways to choose group of 3 formed just by junior partners.

So, the number of possibilities to have at least one junior partner is the sum of the three last options (75 + 60 + 10 = 155).

Finally the probability that at least one of the junior partner is on the committee:

P=155/165  

P= 0.9393

So, the probability that at least one of the committee be a junior partner is 0.9393

stops on an even number.