NDA 2 2024 Maths, question 117

Topic: Probability

Q117

The occurrence of a disease in an industry is such that the workers have $20\%$ chance of suffering from it. What is the probability that out of $6$ workers chosen at random, $4$ or more will suffer from the disease?

(a) $53/3125$ Correct answer
(b) $63/3125$
(c) $73/3125$
(d) $83/3125$

Solution.

Answer: (a) $53/3125$

The number of workers with the disease is binomial with $n = 6$ , $p = \frac{1}{5}$ .
$P(X \geq 4) = \binom{6}{4}\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{2} + \binom{6}{5}\left(\frac{1}{5}\right)^{5}\frac{4}{5} + \left(\frac{1}{5}\right)^{6}$ .
$= \frac{240 + 24 + 1}{15625} = \frac{265}{15625} = \frac{53}{3125}$ .

Trap: The binomial model assumes each worker falls ill independently, which the question takes for granted.