A group consists of 6 men and 5 women. Five people are selected to attend a conference. In how many ways can 5 people be selected from this group of 11? In how many ways can 5 men be selected from the 6 men? What is the probability that the group selected will be all men?

462

6

1.2%

Step-by-step explanation:

Since the questions are combinations, we must apply the combination formula, which is as follows:

n C r = (n!) / (r! (n-r)!)

Because there are 6 men and 5 women, there are a total of 11 people.

Thus:

n = 11

In the first question in how many ways can 5 people be selected from this group of 11, r = 5.

Replacing in the formula:

11 C 5 = (11!) / (5! * (11-5)!)

11 C 5 = (11!) / (5! * 6!)

11 C 5 = 462

In the second question, in how many ways can 5 men be selected from the 6 men, here n = 6 and r = 5, replacing we are left with:

6 C 5 = (6!) / (5! * (6-5)!)

6 C 5 = (6!) / (5! * 1!)

6 C 5 = 6

In the last question of what is the probability that the selected group is all men, we have that it is the combination of the two previous questions. Since the total would be part A it would be the total of the combinations of choosing 5 of 11 people and part B of the 6 men that there are the combinations of choosing 5.

Divide the two values from parts A and B to get ...

(result from part B) / (result from part A) = (# of ways to pick 5 men) / (# of ways to pick 5 people)

(result from part B) / (result from part A) = 6/462

(result from part B) / (result from part A) = 0.012

In other words, the probability is 1.2%

what do u need to know

step-by-step explanation:

step-by-step explanation:

two pairs of adjacent sides are of equal length. this implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. it also implies that the diagonals are perpendicular.