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Question
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution
Let X denote the number of times of getting a ball marked with the digit 0.
P(getting a ball marked with the digit 0) = p = `(1)/(10)`
∴ q = 1 – p = `1 - (1)/(10) = (9)/(10)`
Given, n = 4
∴ X ∼ B`(4, 1/10)`
The p.m.f. of X is given by
P(X = x) = `""^4"C"_x (1/10)^x (9/10)^(4 - x), x` = 0, 1,...,4
P(none is marked with the digit 0) = P(X = 0)
= `""^4"C"_0 (1/10)^0 (9/10)^4`
= `(9/10)^4`.
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