Question

a group consist of four men and five women. three people are selected to attend a conference. a) in how many ways can 3 people be selected from this group of nine? b) in how many ways can three women be selected from the five women? c)find the probability that the selected group will consist of all women

Answer 1

A) \[\frac{ 9\times8\times7\times6\times5\times4\times3\times2\times1 }{ 6\times3\times2\times1 }\]
\[\frac{ 504 }{ 6 } = 84\]

B) \[\frac{ 5\times4\times3\times2\times1 }{ 1\times2\times3\times2 }\]
\[\frac{ 60 }{ 6 }\]

C)You have a 5 in 9 chance of the first person being a woman.
You then have a 4 in 8 chance of the second one being a woman.
Finally, you have a 3 in 7 chance of the third person being a woman.

So the chances are:
\[(\frac{ 5}{ 9 })\times(\frac{ 4}{ 8 })\times(\frac{ 3}{ 7 }) \]
\[\frac{ 5\times4\times3 }{ 9\times8\times7 }\]
\[\frac{ 5 }{ 42 }\]

I hope this helps! XD

Answer 2

thanks it did!

Answer 3

You're welcome! XD