A committee of four is chosen at random from a group of 66 women and 55 men. Find the probability that the committee contains at least one man.

 There are 13 C 4 = 715 ways to choose a group of men and women.

So there are 715 - 70 = 645 ways to choose a group with at least one man.

This means the probability is 645/715 = 129/143 = 0.9020979 (so roughly a 90.20979% chance)

Step-by-step explanation:  There are 13 C 4 = 715 ways to choose a group of men and women.

So there are 715 - 70 = 645 ways to choose a group with at least one man.

This means the probability is 645/715 = 129/143 = 0.9020979 (so roughly a 90.20979% chance)

i need on these three questions. find standard form equation of line perpendicular to y = (2/3)x - 8 that goes through the point

the equation of a line is -6x-2y=-18 what is the x 3 answers sep 21, 2015

finding the critical point for f(x,y)=(x^2)*y+y^3 1 answer mar 10, 2013

more results from answers.yahoo.com

step-by-step explanation:

answer: about 22.67

step-by-step explanation: