SOLUTION: A sample of 4 diffrent calculators is randomly selected from a group containing 18 that are defective and 39 that have no defects. What is the probability that at least one of the
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Question 358945: A sample of 4 diffrent calculators is randomly selected from a group containing 18 that are defective and 39 that have no defects. What is the probability that at least one of the calculators is defective?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
The probability of at least one is the probability of exactly one plus the probability of exactly two plus the probability of exactly three plus the probability of exactly four.
\ =\ \sum_{i=0}^4 \left(4\cr i\right\)\left(\frac{18}{39}\right)^i\left(\frac{21}{39}\right)^{4\,-\,i})
But since that sum plus the probability of exactly zero occurrences equals one, we can calculate the sum by subtracting the probability of exactly zero from 1:
\ =\ 1\ -\ P_4\left(0,\frac{18}{39}\right)\ =\ 1\ -\ \left(4\cr 0\right\)\left(\frac{18}{39}\right)^0\left(\frac{21}{39}\right)^{4})
But since \ =\ 1\ \forall\ n\ \in\ \mathbb{N})
and 
, the above reduces to:
^{4})
You can do the rest of the arithmetic.
John
My calculator said it, I believe it, that settles it
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