Probability Question.Can somone tell me if i am doing right thing

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NOT favour=(6C1)(4C2) + (6C2)(4C1) +(6C3) + (6C2)(7C1)=36+60+20+105=221Sample space - NOT favour = 680-221=459459/Sample Space=Probability of E=459/680=27/40=67.5%IS THIS CORRECT METHOD AND ANSWER?In summary, the probability of a committee vote in favor of a proposal is 36% when all biologists are in favor and everyone else is against, 72.79% when all chemists are against and everyone else is in favor, and 67.5% when all bi
  • #1
cloud360
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I WANNA KNOW IF MY ANSWERS AND METHOD ARE CORRECT

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?
 
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  • #2
For question 1: I like your 245/680.
For question 2: I'm very skeptical about your 515/680 and think that is a little too high and you should check that again.
For question 3: I'm very skeptical about your 459/680 and think that is much too high and you should check that again.
 
  • #3
Bill Simpson said:
For question 1: I like your 245/680.
For question 2: I'm very skeptical about your 515/680 and think that is a little too high and you should check that again.
For question 3: I'm very skeptical about your 459/680 and think that is much too high and you should check that again.
Thanks for your time. I really appreciate it.

Question 1 has 7 people in favour
Question 2 has 11 people in favour (so should be higher than Q1)
Question 3 has 7 people in favour but this time can pass with 1 biologist, given 2 physicists.


Can you tell me what i did in my working out, that made you skeptical.
 
  • #4
I expected my initial hints would be enough. Try again.

For Question 2 I believe neither your (6C2)*(10C1) nor your (6C2)*(10C0) are correct. Why?

For Question 3 I believe your population-opposed is missing something essential. What?

Now, if I am correct, the more important thing for you to discover is why you made this kind of mistakes and why you were not able to discover them and what changes you could make to try to make to avoid this in the future. That will be far more valuable to you than the answers. Too many see an error, quickly scribble in the new answer and completely miss what they could learn from the process.
 
  • #5
Bill Simpson said:
I expected my initial hints would be enough. Try again.

For Question 2 I believe neither your (6C2)*(10C1) nor your (6C2)*(10C0) are correct. Why?

For Question 3 I believe your population-opposed is missing something essential. What?

Now, if I am correct, the more important thing for you to discover is why you made this kind of mistakes and why you were not able to discover them and what changes you could make to try to make to avoid this in the future. That will be far more valuable to you than the answers. Too many see an error, quickly scribble in the new answer and completely miss what they could learn from the process.


do you think i double counted or double subtracted. i.e their is a repeat somewhere, so i need to sibtract something e.t.c


i got an exam tommorow. if you have time. i kindly ask you tell me exactly what i did wrong so i can see the error, as i am kind of worried.

thanks again
 
  • #6
Bill Simpson said:
I expected my initial hints would be enough. Try again.

For Question 2 I believe neither your (6C2)*(10C1) nor your (6C2)*(10C0) are correct. Why?

For Question 3 I believe your population-opposed is missing something essential. What?

Now, if I am correct, the more important thing for you to discover is why you made this kind of mistakes and why you were not able to discover them and what changes you could make to try to make to avoid this in the future. That will be far more valuable to you than the answers. Too many see an error, quickly scribble in the new answer and completely miss what they could learn from the process.

OH i see what i did wrong. it should be (6C2)*(11C1)+(6C2)*(11C0)

thanks for your help
 
  • #7
For question 2
***CORRECTIONS MADE***
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)*(11C1) =165
Event 2= 3 chemist and 0 other=(6C3)*(11C0) =20

NOT E=165+20=185

(Sample space-NOT E) = E=680-185=495.
495/Sample Space=Probability of E=495/680=99/136=72.79%

IS THIS CORRECT METHOD?

For question 3
***CORRECTIONS MADE***
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=(1 chemist + 1 bioligist + 1 physicists)

NOT favour=(6C1)(4C2) + (6C2)(4C1) +(6C3) + (6C2)(7C1) + (6C1)(7C1)(4C1)=36+60+20+105 + 168=389

Sample space - NOT favour = 680-389=291
291/680=42.79%

Final notes
What i have learnt
-Always check if the permutations add up to the sample size (in this case 17)
-Need to learn how a democracy works. If its a tie, then it doesn't pass/no favour

The problems worse for me because i have found in many exams when i see the answer in a calculator, i write another answer down on paper. E.g if i see 1723, i sometimes would write down 1273
 
Last edited:
  • #8
"OH i see what i did wrong. it should be (6C2)*(11C1)+(6C2)*(11C0)"
Still not right.
"Event 1= 2 chemist and 1 other=(6C2)*(11C1) =165
Event 2= 3 chemist and 0 other=(6C3)*(11C0) =20"
Finally right.

"NOT favour=(6C1)(4C2) + (6C2)(4C1) +(6C3) + (6C2)(7C1) + (6C1)(7C1)(4C1)=36+60+20+105 + 168=389"
Still missing one case.

But at least it sounds like you are beginning to recognize the kinds of errors you make. That is a very good thing. Keep watching for the kinds of errors you make and what kinds of circumstances they show up in. And make yourself some rules that you will follow to help yourself catch and correct those mistakes.
 
Last edited:
  • #9
Bill Simpson said:
I expected my initial hints would be enough. Try again.

For Question 2 I believe neither your (6C2)*(10C1) nor your (6C2)*(10C0) are correct. Why?

For Question 3 I believe your population-opposed is missing something essential. What?

Now, if I am correct, the more important thing for you to discover is why you made this kind of mistakes and why you were not able to discover them and what changes you could make to try to make to avoid this in the future. That will be far more valuable to you than the answers. Too many see an error, quickly scribble in the new answer and completely miss what they could learn from the process.

Bill Simpson said:
"OH i see what i did wrong. it should be (6C2)*(11C1)+(6C2)*(11C0)"
Still not right.
"Event 1= 2 chemist and 1 other=(6C2)*(11C1) =165
Event 2= 3 chemist and 0 other=(6C3)*(11C0) =20"
Finally right.

"NOT favour=(6C1)(4C2) + (6C2)(4C1) +(6C3) + (6C2)(7C1) + (6C1)(7C1)(4C1)=36+60+20+105 + 168=389"
Still missing one case.

But at least it sounds like you are beginning to recognize the kinds of errors you make. That is a very good thing. Keep watching for the kinds of errors you make and what kinds of circumstances they show up in. And make yourself some rules that you will follow to help yourself catch and correct those mistakes.

yes, i am missing the case where their is "3 physicists"...i am so grateful for all your help.

you didnt need to take your time to help me, but still did. i can see that you are a decent person. i thank you for all your generosity up to now

thanks again
 
Last edited:
  • #10
For question 3
***CORRECTIONS MADE***
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 + 1 bioligist)
NOT favour=(1 chemist + 1 bioligist + 1 physicists)
NOT favour=(3 physicists)

NOT favour=(6C1)(4C2) + (6C2)(4C1) +(6C3) + (6C2)(7C1) + (6C1)(7C1)(4C1) + (4C3)=36+60+20+105 + 168 + 4=393

Sample space - NOT favour = 680-393=287
287/680=42.21%

Final notes
What i have learnt
-Need to learn how a democracy works (if no one votes, then no favour/no one wins)
 

Related to Probability Question.Can somone tell me if i am doing right thing

1. What is the definition of probability?

Probability is the measure of the likelihood that an event will occur. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

2. How is probability calculated?

Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This can be represented as a fraction, decimal, or percentage.

3. What is the difference between theoretical and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual experiments or observations and may differ from theoretical probability.

4. How is probability used in real life?

Probability is used in many real-life situations, such as weather forecasting, gambling, risk assessment, and business decision-making. It helps us make predictions and evaluate the likelihood of different outcomes.

5. Can probability be greater than 1 or less than 0?

No, probability cannot be greater than 1 or less than 0. These values represent impossibility and certainty, respectively, and are not considered valid probabilities.

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