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Question
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease. Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover
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Solution
n = 5
Let probability of recovery p = 0.73
q = 1 – p = 1 – 0.73
∴ q = 0.27
The binomial distribution is
P(X = x) = nCxpxqn-x
P(X = x) = 5Cx(0.73)x(0.27)5-x
P(atleast 3 of the 5 mice recover) = P(X ≥ 3)
= P(X = 3) + P(X = 4) + P(X = 5)
= 5C3 (0.73)3(0.27)5-3 + 5C4 (0.73)4 (0.27)5-4 + 5C5 (0.73)5 (0.27)5-5
5C2 (0.73)3(0.27)2 + 5c1 (0.73)4 (0.27)1 + 5c0(0.73)5(0.27)°
[`(5 xx 4)/(1 xx 2)` × 0.389017 × 0.0729] + [5 × 0.28398241 × 0.27] + (1 × 0.2073071593 × 1)
= 0.283593393 + 0.3833762535 + 0.2073071593
= 0.2836 + 0.3834 + 0.2073
= 0.8743
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