Probability Homework

lawalkie22 said:

I have a home work question I have been stuck on for a few hours and I cannot figure it out!

"A company selects three computer parts to ship from a supply of 15. Two of the fifteen are known to be defective."
A) What is the probability that none of the parts shipped are defective?
B) What is the probability that at least one of the parts shipped is defective?
C)What is the probability that all of the parts shipped are defective?

For part A, I used P(A)*P(B)*P(C).
2/15= .13333
So,1-.13333=.8667
(.8667)*(.8667)*(.8667)=.6510? No, because probability changes each time

For Part B:
1-.6510(from part A)=.349 Right idea .. if you had A right, this would be the thing to do

Part C:
.1333*.1333*.1333=.0024 No, because probability changes each time

Help?

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DrPhil said:

Thank you for showing your work!

There are two tricks - for parts B and C - so lets look first at part A.

A) P(none) = P(notFirst)*P(notSecond)*P(notThird)
= (13/15)*(12/14)*(11/13) = . . .
After each selection, the total and the number on not-defective each decreases by 1.

B) You already saw the trick! P(at least one) = 1 - P(none)

C) There are only 2 defective, so the probability of getting 3 is zero!

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lawalkie22 said:

Well, I'm glad I at least have the right idea..

So-
A)(13/15)*(12/14)*(11/13)=.6286
B) 1-.6286=.3714
C) So, because I only have 2, the chances of having 3 defective is nearly impossible, which is why the answer would be zero?

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DrPhil said:

C) is really impossible! You can calculate the probability like we did for A):

P(all three) = P(1st)*P(2nd)*P(3rd)
= (2/15)*(1/14)*(0/13) = 0

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lawalkie22 said:

I see.

Is it ok if I ask another problem. Its very similar...

"A company selects three computer parts to ship from a supply of 15. Four of the fifteen are known to be defective"
A) What is the probability that none of the parts shipped are defective?
B) What is the probability that at least one of the parts shipped is defective?
C)What is the probability that all of the parts shipped are defective?


A) (11/15)*(10/14)*(9/13)=.3626
B) 1-.3626=.6374
C) (3/15)*(2/14)*(1/13)= .0022

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DrPhil said:

Yay! Looks great - now I see why they asked for "all" in the first question - didn't make a lot of sense there, but does here.

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