SOLUTION: Suppose that a fisherman has a 0.02 probability of catching a fish on any day he goes out fishing.
What is the probability of him catching at least one fish during a 6 day period?
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Question 998208: Suppose that a fisherman has a 0.02 probability of catching a fish on any day he goes out fishing.
What is the probability of him catching at least one fish during a 6 day period?
Thank you!!
Found 2 solutions by KMST, solver91311:
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
If a fisherman has a probability of catching a fish on any day he goes out fishing,
that fisherman has a probability of not catching amy fish on any day he goes out fishing.
Those would be the fisherman's probability of catching a fish and coming home empty handed each day.
The probability for one day is independent of the probability for another day.
If we consider consecutive days (and even if we considered any 6 random days),
the probability that the fisherman will come home empty handed each one of those days is
(rounded).
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 !})
To find the probability of at least 1, you need 1 minus the probability of Zero. Unless you are a glutton for punishment in which case you can add the probability of 1, plus the probability of 2, plus the probability of 3, and so on through 6.
You want zero successes in 6 trials with a constant probability per trial of 0.02, then subtract from 1.
John
My calculator said it, I believe it, that settles it
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