SOLUTION: basic statistics: A box contains 18 transistors of which 3 are defective. If sample three of the transistors, Find the probability that we would find exactly 1 defective transistor
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Question 281995: basic statistics: A box contains 18 transistors of which 3 are defective. If sample three of the transistors, Find the probability that we would find exactly 1 defective transistor. And find the probability that we would find at least 1 defective transistor.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of selecting a defective transistor on a single trial is 
. The probability of 
successes in 
trials where the probability of success on any given trial is 
is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
where )
is the number of ways to choose things from things and is equal to !})
For the first part of your problem you need to calculate:
\ =\ \left(3\cr 1\right\)\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^{2}\ =\ (3)\left(\frac{1}{6}\right)\left(\frac{25}{36}\right))
There are two ways to do the second part of your problem:
You can calculate
\ =\ P_3(1)\ +\ P_3(2)\ +\ P_3(3))
Which involves considerable drudgery. On the other hand, you could realize that finding at least one is all possibilities EXCEPT finding none, so you could calculate:
\ =\ 1\ -\ P_3(0)\ =\ 1\ -\ \left(3\cr 0\right\)\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{3}\ =\ 1\ -\ (1)(1)\left(\frac{125}{216}\right))
The arithmetic is yours to handle.
John
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