Question 1038517: the probability that a certain make of a car will need repairs in the first six months is 0.35. A dealer sells five such cars. What is the probability that at least one of them will need repairs in the first six months?
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:
\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where  is the number of combinations of things taken at a time and is calculated by !})
You want the probability of "at least 1" success out of 5 trials where the probability of success is 0.35.
If you go at this straight up, you are in for a good deal of nasty arithmetic. You would have to do the above calculation for k = 1, k = 2, k = 3, k = 4, and k = 5 and then add all of the results. But if you realize that the probability of at least 1 is the same thing as 1 minus the probability of zero, then you only have to calculate:
\ =\ 1\ -\ P_5(0,0.35)\ =\ 1\ -\ {{5}\choose{0}}\left(0.35\right)^0\left(0.65\right)^{5})
Then, if you note that  for all positive integers  and that  for all real  , the above reduces to:
\ =\ 1\ -\ P_5(0,0.35)\ =\ 1\ -\ \left(0.65\right)^5)
Just a few key punches on your scientific calculator.
John
My calculator said it, I believe it, that settles it
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