A club is choosing 2 members to serve on a committee. The club has nominated 3 women and 3 men. Based on chance alone, what is the
probability no women are chosen to be on the committee?

The probability that no women are chosen to be in the committee is 2/5 or 0.40.

Step-by-step explanation:

We have two positions, and we have 6 options for those positions, 3 are men, and 3 are women.

If we want to see the probability where no women are chosen, then this means that in the first selection a man is chosen.

If all the 6 nominees have the same probability of being chosen, then the probability that in the first selection a man is chosen is equal to the number of men divided the total number of nominees, this is:

p1 = 3/6 = 1/2.

Now the same happens for the second selection, but now we have 5 total nominees and 2 are men, so the probability now is:

p2 = 2/5

And the joint probability will be the product of the individual probabilities, then we have:

P = p1*p2 = (1/2)*(2/5)  = 2/5.

The probability that no women are chosen to be in the committee is 2/5 or 0.40.

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.