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Question
An experiment succeeds twice as often as it fails, what is the probability that in next five trials there will be at least three successes
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Solution
Success = 2 × fails
p = 2q
⇒ p = 2(1 – p)
p = 2 – 2p
⇒ p + 2p = 2
3p = 2 and p = `2/3`
q = 1 – p = `1 – 2/3`
q = `1/3` and n = 5
The binomial destribution is
P(X = x) = nCxpxqn-x
= `5"C"(2/3)^x (1/3)`
P(atleast three success)
= P(X ≥ 3)
= P(X = 3) + P(X = 4) + P(X = 5)
= `80/243 + 5"c"_4 (2/3)^4 + (1/3)^(5 - 4) + 5"c"_5 (2/5)^5 (1/3)^(5 - 5)`
= `80/243 + 5"C"_1 (16/81)(1/3) + 5"c"_0 (32/243)(1/3)^0`
= `80/243 + 80/243 + 32/243`
= `(80 + 80 + 32)/243`
= `192/243`
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