Question

Normally, 40 % students fail in one examination. Find the probability that at least 4 students in a group of 6 students pass in this examination.

Answer 1

40% student fail in the examination.

∴ Probability that a student fails = \(\frac{40}{100}\) = 0.4

p = Probability that student passes in the examination

= 1 – (Probability that a student fails)

= 1 – 0.4 = 0.6

q = 0.4, n = 6

X = At least 4 students pass in the examination.

∴ X = 4, 5, 6

X is binomial random variable.

∴ P(X = x) = p(x) = ⁿC_x p^x q^n-x

Putting, n = 6, p = 0.6, q = 0.4

P(X = x) = p(x) = ⁶C_x (0.6)^x (0.4)^6-x

∴ p(4) = ⁶C₄ (0.6)⁴ (0.4)^6-4

= \(\left (\frac{6×5×4!}{2!×4!}\right)\) (0.1296) (0.4)²

= (15) (0.1296) (0.16) = 0.3110

∴ p(5) = ⁶C₅ (0.6)⁵ (0.4)^6-5

= \(\left (\frac{6×5!}{1×5!}\right )\) (0.0778) (0.4)

= (6) (0.0778) (0.4) = 0.1866p(6) = ⁶C₆ (0.6)⁶ (0.4)^6-6
= (1) (0.0467) (1) = 0.0467

Now, P(X ≥ 4) = p(4) + p(5) + p(6)
= 0.3110 + 0.1866 + 0.0467
= 0.5443

Hence, the probability that at least 4 students pass in the examination obtained is 0.5443.