Question 919954: A group of 2 women and 5 men decide to send a delegation to their supervisor to discuss grievances.
If a delegation of 2 people is selected at random, determine the probability that at least 1 woman is on the delegation.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there are 2 women and 5 men.
the probability that there will be at least one woman on the delegation is equal to 1 minus the probability that both people on the delegation will be men.
2 men on the delegation means no women.
not 2 men on the delegation means at least one woman.
the probability that there will be 2 men on the delegation is equal to 5/7 * 4/6 = 20/42 which is equal to 10/21.
1 - 10/21 = 11/21
the probability that there will be at least one woman on the delegation is therefore equal to 11/21.
this can also be calculated as follows:
the number of ways you can get 2 men from 5 men without order is equal to c(5,2) which is equal to 5! / (2! * 3!) which is equal to (5*4*3!) / (2!*3!) which is equal to (5*4)/2! which is equal to 20/2 which is equal to 10.
the number of ways you can get 2 people from 7 people without order is equal to c(7,2) which is equal to 7! / (2! * 5!) which is equal to (7*6*5!) / (2!*5!) which is equal to (7*6) / 2! which is equal to 42/2 which is equal to 21.
the probability that both will be men is equal to the number of ways you can get 2 men from 5 men divided by the number of ways you can get 2 people from 7 people which is equal to 10/21.
1 minus that is equal to 11/21 which is the number of ways you can get at least 1 woman on the delegation divided by the number of ways you can get 2 people on the delegation.
if you don't get 2 men on the delegation, then at least one of the people on the delegation has to be a woman.
your solution is 11/21.
this is the probability that at least one member of the delegation will be a woman.
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