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Question
Hospital records show that of patients suffering from a certain disease 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?
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Solution
Let x be the random variable of patience suffering from a certain disease
In a binomial distribution
P(X = x) = ncxpxqn-x
P(X = 4) = `""^6"C"_4 (1/4)^4 (3/4)^(6 - 4)`
= `""^6"C"_2 (1/4)^4 (3/4)^2`
= `(6 xx 5)/(1 xx 2) [9/(4)^6]`
= `15 xx [9/40946]`
= `135/4096`
= 0.03295
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