There are 14 boys and 6 girls giving interview for a job. If three of them are selected, then what is the probability that one of the three is a girl and the other two are the boys?
Total Number of candidates = 20 Let ‘S’ be the Sample Space Then, n(S) = number of ways of three candidates who got selected n(S) = 20C3 = 1140 Let ‘E’ be the event of 1 girl and 2 boys selected Therefore, n(E) = number of possibility of 1 girl out of 6 and 2 boys out of 14 n(E) = 6C1 × 14C2 = 546 Now the required Probability = (n(E))/(n(S)) = 546/1140 = 91/190
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