A committe of 4 is to be chossen from 3 boys(Jason,Carlos,And Bill) and 4 girls. What's the probability a committee will contain Jason.?
2007-11-06 09:38:24 · 7 answers · asked by helppmee =] 2 in Science & Mathematics ➔ Mathematics
Total number of people to choose the committee from: 3+4=7
Number of people to choose the committee from (excluding Jason): 7-1 = 6
No. of ways a committee can be chosen: 7C4 = 35
No. of ways a committee can be chosen WITHOUT Jason: 6C4 = 15
There are 35-15=20 ways to choose a committee INCLUDING Jason.
The probability that the committee will include Jason is 20/35 = 4/7
2007-11-06 09:46:00 · answer #1 · answered by gudspeling 7 · 0⤊ 0⤋
The probability is 4/7 or about 57%.
The number of boys and girls don't matter. Just add them up: 7 people total.
The answer will be (the number of ways Jason can be in the committee) divided by (the total number of possible committees).
Let's find the number of ways Jason can be in the committee first. One of them the members has to be Jason, of course. The other three com from the remaining six people. How many ways are there to pick 3 from 6 (not counting order)? In notation it's 6C3 or C(6,3). It's
6!/[(6-3)!3!] which is
6!/(3!3!) which is
6*5*4/6 which is
20
There are 20 ways to make a committee with Jason in it.
Next we find the total number of ways to make a committee. That's 7C4 or C(7,4).
7! / [ (7-4)!4! ] =
7! / (3!4!) =
7*6*5/6 =
7*5 =
35
So there are 35 ways to make any committee.
Now, put 20 over 35:
20/35 = 4/7 = 57%
So, the probability that Jason will be chosen for the commitee is 4/7 or 57%.
2007-11-06 09:54:29 · answer #2 · answered by ultimatelyconfused 2 · 0⤊ 0⤋
ways to choose Jason, 1
ways to choose 3 people not Jason from 6 people, 6C3 = (6•5•4)/(1•2•3) = 20
ways to choose 4 people from 7, 7C4 = (7•6•5•4)/(1•2•3•4) = 35
so 1(20)/35 = 4/7
2007-11-06 09:43:50 · answer #3 · answered by Philo 7 · 0⤊ 0⤋
You pick 3, besides Jason, out of the other 6, and compare it with the number you pick 4 out of 7.
p = 6C3/7C4 = 4/7
2007-11-06 09:52:10 · answer #4 · answered by sahsjing 7 · 0⤊ 0⤋
There are 4 people chosen, so Jason has 4 chances out of 7 people.
4/7
2007-11-06 09:43:06 · answer #5 · answered by brett s 2 · 1⤊ 0⤋
the total possiblity is 7C4 = 35
number of ways that jason is included is:
(1 * 6 * 5 * 4 ) / 3! = 20
P = 20/35
P = 4/7 <== answer
2007-11-06 09:49:34 · answer #6 · answered by Anonymous · 0⤊ 0⤋
P(jason in)
= 6C3 / 7C4
= 20 / 35
= 4/7
= 57.143%
2007-11-08 07:33:24 · answer #7 · answered by Mugen is Strong 7 · 0⤊ 0⤋