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cedessam123 · Apr 20, 2009, 08:03 PM

A sample of 4 different calculators is randomly selected from a group containing 15 that are defective and 30 that have no defects. What is the probability that at least one of the calculators is defective?:confused:

ROLCAM · Apr 20, 2009, 08:32 PM

I think that the calculation is as follows:-

45*44*43*42 / 4*3*2*1
= 15*11*43*21
= 148,995 to one

Unknown008 · Apr 21, 2009, 04:22 AM

Basically, you have to calculate the probability of having all non defective. Then, subtract that probability from 1, that's your answer!

Here, you are eliminating all the cases where you have all non defective calculators. The remaining possibilities include 1, 2 or 3 non defective calculators.

galactus · Apr 21, 2009, 05:38 AM

As Unknown said, The best way to handle an "at least one" problem to to find the probability of none and subtract from 1.

In order to have no defects, we select 4 from the no defect pile and 0 from the defect pile.

$1-\frac{C(30, 4)\cdot C(15,0)}{C(45,4)}$

Another way:

$1-(\frac{30}{45})(\frac{29}{44})(\frac{28}{43})(\fra c{27}{42})$

Sometimes the LaTex displays wacky even when the code is OK. That last one should be 27/42.

Unknown008 · Apr 21, 2009, 10:20 AM

Originally Posted by galactus

As Unknown said, The best way to handle an "at least one" problem to to find the probability of none and subtract from 1.

In order to have no defects, we select 4 from the no defect pile and 0 from the defect pile.

$1-\frac{C(30, 4)\cdot C(15,0)}{C(45,4)}$

Another way:

$1-(\frac{30}{45})(\frac{29}{44})(\frac{28}{43})(\fra c{27}{42})$

Sometimes the LaTex displays wacky even when the code is OK. That last one should be 27/42.

Ok, here it is:

$1-(\frac{30}{45})(\frac{29}{44})(\frac{28}{43}) (\frac{27}{42})$

Just put a space, not that it changes anything in the display but it corrects the 'wacky'.