The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

P(X) =  $\frac{1}{5}$, P(Y) = $\frac{1}{4}$ , P(Z) =  $\frac{1}{3}$

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

= $\frac{1}{4}*\frac{1}{3}*\frac{4}{5}+\frac{3}{4}*\frac{1}{3}*\frac{1}{5}+\frac{2}{3}*\frac{1}{4}*\frac{1}{5}+\frac{1}{4}*\frac{1}{3}*\frac{1}{5}$

=  $\frac{4}{60}+\frac{3}{60}+\frac{2}{60}+\frac{1}{60}$=  $\frac{10}{60}$=  $\frac{1}{6}$