What is the probability of choosing at least one defective calculator?

[Azrioch]

Hey...

Well, I've got this problem which just is confusing me greatly.

"A sample of 4 different calculators is randomly selected from a group containing 15 that are defective and 33 that have no defects. What is the probability that at leastone of the calculators is defective?"

Now, we're in the binomial probability problems section so I think of the criteria and see that it fails that the events are mutually exclusive so it cannot be that.

I don't have the answer and really just need the method for how to solve it. Crunching numbers is the easy part.

I think P(>=1) = P(X=1 or X=2 or X=3 or X=4)
= P(X=1) + P(X=2) + P(X=3) + P(X=4) - P(1 and 2 and 3 and 4) which would be the addition rule, but I have no idea.

For those probabilities I would do 1/48, 1/47, 1/46, 1/45 and ... no idea for the last one. All that we've done is addition rule with two probabilities.

Im very lost.

Any ideas?

 

[master_coda]

A much easier approach would be to find the probability that none of the calculators are defective, then subtract that from 1 to get that probability that at least one is defective.

 

[Azrioch]

Huhmm.

P(None are defective) = 33/48 + 32/47 + 31/46 + 30/45

P(Atleast one is defective) = 1 - (that sum)

I know that the compliment is right, but is the calculation of the P(None are defective) correct?

Edit:

Mm.. no that can't be right.
It comes out to greater than 1.

 

[Azrioch]

Actually, come to think of it...

Those are supposed to be multiplied, not added, correct?

If so and using those numbers, the probability is 0.21 and the compliment, that is, the chance of at least one calculator being defective is 0.79.

That'd just be an extended multiplication rule (P(E and F) = P(E)*P(F|E)).

Confirmation?

 

[master_coda]

Actually, come to think of it...

Those are supposed to be multiplied, not added, correct?

If so and using those numbers, the probability is 0.21 and the compliment, that is, the chance of at least one calculator being defective is 0.79.

That'd just be an extended multiplication rule (P(E and F) = P(E)*P(F|E)).

Confirmation?

Yes, multiplication is what you want to do, not addition. Your answer now seems correct.

 

[adil]

Here is a method for solving the problem. As the person above stated, take the probability that no defects are chosen and subtract that from 1. The probability that no defects are chosen is (33 choose 4) (choosing 4 nondefectives) divided by the #of ways to choose 4 from the total collection which is (48 choose 4). Whatever that quotient is can then be subtracted from 1. I believe this is the correct way to solve the problem.

 

[master_coda]

Here is a method for solving the problem. As the person above stated, take the probability that no defects are chosen and subtract that from 1. The probability that no defects are chosen is (33 choose 4) (choosing 4 nondefectives) divided by the #of ways to choose 4 from the total collection which is (48 choose 4). Whatever that quotient is can then be subtracted from 1. I believe this is the correct way to solve the problem.

That's another way to find the probability of no defects chosen; it's the one I used to check Azrioch's answer.