committee of 4 persons is to be selected from 8 men to 10 women. Find the probability that at least 2 are men.

Question

PROBABILITY PROBLEM

Answer 1

the answer from blighmas... is correct. Because the next two answers are not, I'll post another to remove ambiguity and explain a little better in case you don't know how it's arrived at.

The number of ways to choose x individuals from a group n is given by (n!)/(x!(n-x)!)

if you've been asked this question it's safe to assume you know how to use factorials...

so the easiest way to figure this probability is to consider all possible outcomes (of equal probability, assuming the boss is fair).

there are 18!/(4!14!) = 3060 ways to pick 4 employees.

since the cases of 0 men and 1 man are easiest to calculate, work with P(at least 2) = 1 - P(0 or 1)

the number of outcomes that have 0 men is easy enough, because every woman can be picked in each of those outcomes, so it's simply 10 choose 4, or 10!/(6!*4!) or 210.

the # of outcomes with 1 man will include every way of picking 3 of 10 women, and every way of picking 1 of 8 men. that's (10 choose 3) * (8 choose 1) or
10!/3!7!) * 8!/(1!7!) = 120*8 = 960.

thus of 3060 outcomes of equal probability, 960 + 210 = 1170 will have less than 2 men. so
P(at least 2) = 1 - 1170/3060 = 1890/3060

=63/102 = 21/34.

As for bach's answer below, he sets up his problem almost arbitrarily. 8 choose 4 is the number of ways to pick 4 men from 8, and represents the number of possible outcomes in which all 4 committee members are men. Thus 8C4/18C4 is simply the probability that all committee members are men, and 1-8C4/18C4 is the probability of at least one woman on the committee (not all men). The expression has little to do with event A as bach defined it. In addition, bach incorrectly calculated 18C4 as 6120. It is 3060. As if that weren't enough he then bungles the arithmetic... (1-70/6120 = 0.988, not .011) Not to mention the never-defined "event B." It's too bad you've had to sift through so many wrong answers here.

Answer 2

Out of total 18 person , a committee of 4 can be selected as 18C4 = (18 * 17 * 16 * 15)/ (4 * 3 * 2 * 1) = 3060 . This is equal to the number of sample spaces.And

The number of favourable events of having atleast 2 men will be selected in either of the following way ,

(1) 2 men and 2 women = 8C2 * 10C2 = 28 * 45 = 1260 or
(2) 3 men and 1 women = 8C3 * 10C1 = 56 * 10 = 560 or
(3) 4 men = 8C4 = 112

Therefore total no. of favourable cases = 1260 + 560 + 112 = 1932

Therefore the required probability = 1932 / 3060 = 161/255

Answer 3

Please note that this is just one way of solving it (and pretty lengthy at that). If the selection of candidates is totally random, then there are these possibilities (notice that the order doesn't really matter):

w, w, w, w
w, w, w, m
w, w, m, w
w, w, m, m **
w, m, w, w
w, m, w, m **
w, m, m, w **
w, m, m, m
m, w, w, w
m, w, w, m **
m, w, m, w **
m, w, m, m
m, m, w, w **
m, m, w, m
m, m, m, w
m, m, m, m

There are 5 unique group combos (try to come to that number without writing down all the combinations), and these are:

w, w, w, w (7 / 306)
w, w, w, m (7 / 153)
w, w, m, m (7 / 102) (6 of these - check the upper list)
w, m, m, m (4 / 51) (4 of these - same as above)
m, m, m, m (7 / 102)

The odds we are looking for are calculated as follows:

6 * 7 / 102 + 4 * 4 / 51 + 7 / 102 = 81 / 102 = 27 / 34

Why like this? Because, if you were to write down the odds of all the sixteen possibilities, and add them up, you would get 1 (or 306 / 306). If you pick out those that have two or more men in them, you would get 243 / 306 = 81 / 102 = 27 / 34.

Answer 4

Total possible for any 4 = 18!/14!4!

Let's figure possible combinations that include LESS than 2 men:

Exactly 1 man = 8*(10!/7!3!)
No men = 10!/6!4!

18!/14!4! = 18*17*16*15/24= 12*15*17=3060
8*10!/7!3!= 8*10*9*8/6=8*10*12=960
10!/6!4!= 10*9*8*7/24=10*3*7=210

Total possible combinations (any) - 1 man only - no men =

3060-960-210=1890
Probability = 1890/3060=21/34

BTW, miinii is wrong - that is calculation for EXACTLY 2 men - not AT LEAST 2 men

Answer 5

From 18 persons 4 persons can be selected in 18C4 ways. Thus there are 18C4 equally likely, mutually, exclusive and exhaustive outcomes.
Let events A and B be
A: at least 2 of the selected persons are men. i.e. 4 men can be selected out of 8 men in 8C4 ways.
P(at least 2are men) = 1 - 8C4/18C4
= 1 - 70/6120
= .0113