Arithmetic Ability
 Probability

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Q.

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.

View Answer

Correct choice: C

Explanation:

P(X) = 1/5, P(Y) =1/4 , P(Z) = 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) =(1/4*1/3*4/5)+(3/4*1/3*1/5)+(2/3*1/4*1/5)+(1/4*1/3*1/5) = 4/60+3/60+2/60+1/60 = 10/60= 1/6
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