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Question
If the probability of success is 0.09, how many trials are needed to have a probability of atleast one success as 1/3 or more?
Solution
Given p = 0.09 ......(success)
q = 0.91 .......(failure)
We have to find number of trials ‘n.’
According to the problem,
P(X ≥ 1 ) > `1/3`
We must have atleast one success
1 – P(X < 1) > `1/3`
1 – P(X = 0) > `1/3`
or
P(X = 0) < `2/3`
Using p.m.f, we have,
`""^"n""C"_0 (0.09)^0 (0.91)^"n" < 2/3`
`(0.91)"n" < 2/3`
We can use log tables to calculate or by trial method try for n = 1, 2,…… using calculator.
We observe that `(0.91)5 < 2/3`.
Thus we need minimum 5 trial or more.
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