Question

Three men and three women are to be randomly seated in a row. Find the probability that both the end places will be filled by women.

Answer 1

i think its 1/5

Answer 2

Like most probability problems, you need to figure out how many possibilities there are total (sample space), and how many ways among them can you get the result you're interested in.

For this problem, the total possibilities is all the ways the 6 people can sit:
6 choices for the first seat, then 5 for the next, 4 for the next, and so on, so
6x5x4x3x2x1=6!=720.

Now we need the number of ways that 2 women can sit in the ends, which looks like this:
W_ _ _ _ W, so filling in seats with possible choices gives Wx4x3x2x1xW, or 4!=24.

Lastly, we need to figure out the number of ways to pick which two women sit on the ends, which is again 3 choices for the one end, then 2 choices for the other, so 3x2 = 6.

The probability then is total ways for women to sit on the ends, 6x24 = 144, divided by the total ways for people to sit = 720. So, 144/720 = 1/5.

Answer 3

I think it can be simpler. We pick at random to fill the first end seat (3/6 are women). Now we pick at random for the second end seat (2/5 are women). The probability for both events is 3/6 times 2/5 = 1/5. Same answer.