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.
Solution
Correct Answer: Option C
P(X) = \(\frac15\) , P(Y) = \(\frac14\) , P(Z) = \(\frac13\)
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)
= \(\left(\frac14\times\frac13\times\frac45\right)+\left(\frac34\times\frac13\times\frac15\right)+\left(\frac23\times\frac14\times\frac15\right)+\left(\frac14\times\frac13\times\frac15\right)\)
= \(\frac4{60}+\frac{\displaystyle3}{\displaystyle60}+\frac{\displaystyle2}{\displaystyle60}+\frac{\displaystyle1}{\displaystyle60}\)
= \(\frac{10}{60}\)
= \(\frac16\)
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