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I WANNA KNOW IF MY ANSWERS AND METHOD ARE CORRECT
A science committee of 3 people is to be formed from a group of 17 scientists (7 biologists, 4
physicists and 6 chemists) with each possible committee equally likely.
What is the probability the committee vote in favour of a proposal given:-
(a) all biologists in favour, everybody else against?
(b) all chemists against, everybody else in favour?
(c) all biologists in favour, all chemists against and all physicists abstain?
none
Favour =Majority of comitte favour i.e at least 2 of 3 favour
For question 1
Sample space = (17C3)=680
If biologist all favour.. Then for favour, 2 biologist or 3 biologist in the comitee
Event 1= 2 biologist and 1 other=(7C2)*(10C1) =210
Event 2= 3 biologist and 0 other=(7C2)*(10C0) =35
245/Sample space = 245/680 = 49/136=36%
IS THIS CORRECT?
For question 2
Sample space = (17C3)=680
If bioligist all favour + phyicists all favour + all chemist against. Then we have many events. So is better to use addition rule.
Addition rule = E + NOT E=Sample Space.....where E= favour
NOT E= 2 chemists or 3 chemists in comitee
Event 1= 2 chemist and 1 other=(6C2)*(10C1) =150
Event 2= 3 chemist and 0 other=(6C2)*(10C0) =15
NOT E=150+15=165
(Sample space-NOT E) = E=680-165=515
515/Sample Space=Probability of E=515/680=103/136=75.7%
IS THIS CORRECT METHOD?
For question 3
Sample space = (17C3)=680
If all biologist favour + all chemist against + all physicists don't give a damn.
Favour = (3 biologist)+(2 biologist+1 Other) =Question 1
Favour = (1 bioligist + 2 physicists) +(2 bioligist + 1 physicists)
Favour = (2 bioligist + 1 chemist)
Then we have many events. So is better to use addition rule.
NOT favour=(1 chemist + 2 physicist) + (2 chemist + 1 physicist) + (3 chemist)
NOT favour=(2 chemist + bioligist)
NOT favour=(6C1)(4C2) + (6C2)(4C1) +(6C3) + (6C2)(7C1)=36+60+20+105=221
Sample space - NOT favour = 680-221=459
459/680=27/40=67.5%
IS THIS CORRECT METHOD AND ANSWER?
Homework Statement
A science committee of 3 people is to be formed from a group of 17 scientists (7 biologists, 4
physicists and 6 chemists) with each possible committee equally likely.
What is the probability the committee vote in favour of a proposal given:-
(a) all biologists in favour, everybody else against?
(b) all chemists against, everybody else in favour?
(c) all biologists in favour, all chemists against and all physicists abstain?
Homework Equations
none
The Attempt at a Solution
Favour =Majority of comitte favour i.e at least 2 of 3 favour
For question 1
Sample space = (17C3)=680
If biologist all favour.. Then for favour, 2 biologist or 3 biologist in the comitee
Event 1= 2 biologist and 1 other=(7C2)*(10C1) =210
Event 2= 3 biologist and 0 other=(7C2)*(10C0) =35
245/Sample space = 245/680 = 49/136=36%
IS THIS CORRECT?
For question 2
Sample space = (17C3)=680
If bioligist all favour + phyicists all favour + all chemist against. Then we have many events. So is better to use addition rule.
Addition rule = E + NOT E=Sample Space.....where E= favour
NOT E= 2 chemists or 3 chemists in comitee
Event 1= 2 chemist and 1 other=(6C2)*(10C1) =150
Event 2= 3 chemist and 0 other=(6C2)*(10C0) =15
NOT E=150+15=165
(Sample space-NOT E) = E=680-165=515
515/Sample Space=Probability of E=515/680=103/136=75.7%
IS THIS CORRECT METHOD?
For question 3
Sample space = (17C3)=680
If all biologist favour + all chemist against + all physicists don't give a damn.
Favour = (3 biologist)+(2 biologist+1 Other) =Question 1
Favour = (1 bioligist + 2 physicists) +(2 bioligist + 1 physicists)
Favour = (2 bioligist + 1 chemist)
Then we have many events. So is better to use addition rule.
NOT favour=(1 chemist + 2 physicist) + (2 chemist + 1 physicist) + (3 chemist)
NOT favour=(2 chemist + bioligist)
NOT favour=(6C1)(4C2) + (6C2)(4C1) +(6C3) + (6C2)(7C1)=36+60+20+105=221
Sample space - NOT favour = 680-221=459
459/680=27/40=67.5%
IS THIS CORRECT METHOD AND ANSWER?