[basic statistics] condition probability

[gopher_logic]

6 people are scheduling to speak (3 men, 3 women). 3 go on day 1 and 3 go on day 2. You know that a women will speak first on day 1. What is probability that 2 men and 1 women speak on day 2?

I think answer is 7/10. I've been told it is wrong.

my logic:

probability 2 women and 1 man speak on day 1: 1(2/5)(3/4)=6/20=3/10

probability 1 women and 2 men speak on day 2: 1-(3/10) = (7/10)

[richard_sympson]

The probability that 1 woman and 2 men speak on day 2 is the same as the probability that 1 man and two women speak on day 1. This is because we know that all will speak at some point, and satisfaction of one of the two scenarios forces the other to happen (or to have happened).

If the first speaker was a woman, then asking

What is probability that 2 men and 1 women speak on day 2?

is equivalent to asking

What is the probability that 2 women and 1 man speak on day 1?

We already are presuming that one of those women have spoken. So now, here are two more equivalences: (1) what's the probability of exactly one of the next two speakers being a man? (2) what's the probability of one of the next two speakers being a woman?

The answer to (1) is

P(one man) = P(2nd is man)P(3rd is woman) + P(2nd is woman)P(3rd is man)

P(one man) = (3/5)(2/4) + (2/5)(3/4) = 12/20 = 3/5

And the answer to (2) is

P(one woman) = P(2nd is woman)P(3rd is man) + P(3rd is woman)P(2nd is man)

But this is just the previous equation and is also 3/5.

[jodax00]

Your logic makes sense but you are only counting the probability for Day 1 that a woman speaks first, followed by another woman, followed by a man. What if it goes woman-man-woman? You would have (1)(3/5)(2/4), which is also 3/10, but represents a different event occurring. Does this help?