This is an example question, but I'm not sure how my instructor came up with the answer. Could anyone explain in detail please.

Among the contestants in a competition are 49 women and 29 men. If 5 winners are randomly selected, what is the probability that they are all men.

The probablity of the first man is 29/78 where 29 is the number men available, and 78 is the total population
The probabliity of the second man given that one has been selected is ..
28/77

The joint probability of five is ..

29/78 * 28/77 * 27/76 * 26/75 * 25/74

If you have any questions on this, reply.

Thank you so much. I totally get it now. I have another problem like that but with 36 women and 28 men. 5 chosen randomly with the probability that they're all men. I found the answer P= .01289 .Thanks you really helped!

all men? (28)/62 * 27/61 * 26/60 and so forth.

isn't it suppose to be 28/64 where 64 is the total population of men and women?

Of course, I added wrong in my head. I have to stop doing that at my age. Thanks for reading it carefully, and finding the error. Good work.

You're welcome! I'm glad I could help clarify that for you. It's important to keep track of the total population accurately when calculating probabilities.

For the second problem, you are correct that the total population is 64 (36 women + 28 men). So, the probability of selecting the first man would be 28/64. The probability of selecting the second man, given that a man has already been selected, would be 27/63. And so on.

The joint probability of selecting 5 men can be calculated as:

(28/64) * (27/63) * (26/62) * (25/61) * (24/60)

Which equals approximately 0.01289, as you mentioned.

Great job on your calculations, and keep up the good work! If you have any further questions, feel free to ask.