3 students appear at an examination of mathematics . the probability of their success are 1 / 2 , 1 / 4 , 1 / 5 respectively . find the probability of success of at least two .
the probability of success of at least two students will involve the following possibilities . the first two students are successful , the last two students are successful , the first and third students are successful and all the three students are successful . therefore , the required probability = 1 / 2 x 1 / 4 x 4 / 5 + 1 / 4 x 1 / 5 x 1 / 2 + 1 / 2 x 1 / 5 x 3 / 4 + 1 / 2 x 1 / 4 x 1 / 5 = 9 / 40 answer : c
add(multiply(multiply(divide(1, 2), divide(1, 4)), divide(1, 5)), add(add(multiply(multiply(divide(1, 2), divide(1, 5)), subtract(1, divide(1, 4))), multiply(divide(1, 2), divide(1, 5))), multiply(multiply(divide(1, 4), divide(1, 5)), subtract(1, divide(1, 2)))))