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

Answer 1

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) = nC_xp^xq^n-x

P(X = x) = 5C_x(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