Question

Three students appear at an examination of mathematics. The probability of their success are 1/3 , 1/4 , 1/5 respectively. Find the probability of success of at least two.

Answer 1

1/6

Answer 2

Let $p_1, p_2, p_3$ be the probabilities of success for the three students, and $q_1, q_2, q_3$ be the probabilities of failure.

Given: $p_1 = \frac{1}{3}$, $p_2 = \frac{1}{4}$, $p_3 = \frac{1}{5}$

Probabilities of failure: $q_1 = 1 - \frac{1}{3} = \frac{2}{3}$ $q_2 = 1 - \frac{1}{4} = \frac{3}{4}$ $q_3 = 1 - \frac{1}{5} = \frac{4}{5}$

Probability of exactly two successes: $p_1 p_2 q_3 + p_1 q_2 p_3 + q_1 p_2 p_3 = (\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}) + (\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5})$ $= \frac{4}{60} + \frac{3}{60} + \frac{2}{60} = \frac{9}{60}$

Probability of exactly three successes: $p_1 p_2 p_3 = \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} = \frac{1}{60}$

Probability of at least two successes = P(exactly 2 successes) + P(exactly 3 successes) $= \frac{9}{60} + \frac{1}{60} = \frac{10}{60} = \frac{1}{6}$