SOLUTION: A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys? I know th

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Question 287482: A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
I know the answer is 59/143 = .413
I don't know the steps of how to solve the problem.
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
I know the answer is 59/143 = .413
I don't know the steps of how to solve the problem.
---------------------------------------------------------
P(3<= x <=5) = p(x=3) + P(x=4) + P(x=5)
= 6C3/13C3 + 6C4/13C3 + 6C5/13C3
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= [6C3+6C4+6C5]/13C3
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= [20+15+6]/286
---
= 41/286
= 0.1434
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Cheers,
Stan H.
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Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys? 


The other tutor's solution is wrong.

We get the probability of the complement event
and subtract from 1.  The complement event is a committee with
no boys or 1 boy or 2 boys.

1 - P(0 boys OR 1 boy OR 2 boys)

"OR" means "ADD"

1 - [P(0 boys)+P(1 boy)+P(2 boys)]

P(0 boys) = P(5 girls) = 

P(1 boy) =  P(4 girls AND 1 boy) 

"AND" means "MULTIPLY", so

P(1 boy) =  P(4 girls AND 1 boy) = 

P(2 boys) =  P(3 girls AND 2 boys) = 

So,

1 - [P(0 boys)+P(1 boy)+P(2 boys)]

becomes:

1 - []=

Edwin
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