probabilty (selecting committee members from group)

[tsh44]

Hello I found the answers to this problem one way but I would like to know how to solve it in another easier way using combinations.

A committee of 3 people is to be randomly selceted from the 6 people Archibald (A), Beatrix (B), Charlene (C), Denise (D), Eloise (E) and Fernado (F).

Find the probability that Eloise and Fernando are on the same committee
Find the probability that neither Elosie nor Fernando is on the committee
Find the probability that Denise is on the committee given that neither Archibald nor Beatrix is.

I used a sample space and simply counted. My answers were 1/5, 1/5, and 1/10 .
However I also know that you can use cominations to do such a problem so you do not have to write out a sample space if the problem is too large. However I am not sure what numbers I would use in each case for the nominator and denominator.. Thank you for any help.

To show you wht I mean I wrote out this problem in the book.

Five cards are drawn at random form a standard deck. Find th eprobabilty that all 5 cards are hearts.

P = 13 C 5 / 52 C 5 = .274

 

[pka]

Re: probabilty

A committee of 3 people is to be randomly selceted from the 6 people Archibald (A), Beatrix (B), Charlene (C), Denise (D), Eloise (E) and Fernado (F).
Find the probability that Eloise and Fernando are on the same committee
Find the probability that neither Elosie nor Fernando is on the committee
Find the probability that Denise is on the committee given that neither Archibald nor Beatrix is.

Five cards are drawn at random form a standard deck. Find th eprobabilty that all 5 cards are hearts.
P = 13 C 5 / 52 C 5 = .274 [CORRECT!]

 

[soroban]

Hello, tsh44!

A committee of 3 people is to be randomly selected from the 6 people:
Archibald (A), Beatrix (B), Charlene (C), Denise (D), Eloise (E) and Fernando (F).

(1) Find the probability that E and F are on the same committee
(2) Find the probability that neither E nor F is on the committee
(3) Find the probability that D is on the committee given that neither A nor B is.

There are: \(\displaystyle \,_6C_3\:=\:\frac{6!}{3!3!} \:= \:20\) possible committees.


(1) Since \(\displaystyle E\) and \(\displaystyle F\) are on the committee, there are 4 choices for the third member.

Therefore: \(\displaystyle \,P(E\text{ and }F)\:=\:\frac{4}{20} \:= \:\frac{1}{5}\) . . . You are correct!


(2) If \(\displaystyle E\) and \(\displaystyle F\) are not on the committee, the 3 members are chosen from the other 4 people.

\(\displaystyle \;\;\)There are: \(\displaystyle \,_4C_3\:=\:\frac{4!}{3!1!}\:=\:4\) ways.

Therefore: \(\displaystyle \,P(\text{neither }E\text{ nor }F)\:=\:\frac{4}{20}\:=\:\frac{1}{5}\) . . . You are right!


(3) Since neither \(\displaystyle A\) nor \(\displaystyle B\) is on the committee, only \(\displaystyle \{C,\,D,\,E,\,F\}\) are available.
There are are: \(\displaystyle \,_4C_3\:=\:\frac{4!}{3!1!}\:=\;4\) possible committees.

Since \(\displaystyle D\) is on the committee, the other two are chosen from the remaining three people.
There are: \(\displaystyle \,_3C_2\:=\:\frac{3!}{2!1!} \:=\:3\) ways.

Therefore: \(\displaystyle \,P(D\,|\text{ not }A\text{ and not }B}) \:=\:\frac{3}{4}\)


Your sample space must have been incorrect.

Since \(\displaystyle A\) and \(\displaystyle B\) are not on the committee,
\(\displaystyle \;\;\)the possible committees are : \(\displaystyle \,(C,D,E),\;(C,D,F),\;(C,E,F),\;(D,E,F)\)

And \(\displaystyle D\) is on the committee in three of the cases.

 

[tsh44]

thank you pka