Web-Probability

Question

A candidate is selected for interview of management trainees for $3$ companies.

For the first company, there are $12$ candidates, for the second there are $15$ candidates and for the third, there are $10$ candidates. Find the probability that he is selected in at least one of the companies.

Assume that $1$ candidate will be selected in each of the interviews, and all candidates appearing for the interview have an equal probability of getting selected.

Answer 1

1- not selected in any company = selected in at least one of the company.

1-($\frac{11}{12}*\frac{14}{15}*\frac{9}{10}$)= 0.23

 

another approach is selected in only 1 + selected in 2 + selected in all

= $\frac{1}{12}*\frac{14}{15}*\frac{9}{10}+\frac{11}{12}*\frac{1}{14}*\frac{9}{10}+\frac{11}{12}*\frac{14}{15}*\frac{1}{10}$

+$\frac{1}{12}*\frac{1}{15}*\frac{9}{10}+\frac{1}{12}*\frac{14}{15}*\frac{1}{10}+\frac{11}{12}*\frac{1}{15}*\frac{1}{10}$

+$\frac{1}{12}*\frac{1}{15}*\frac{1}{10}$=0.23

Comments

@ Tesla! Sir, Are we not considering companies probability here ?

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It doesn't make any sense

Answer 2

So to get selected in at least one company we have various combination i.e get selected in 1st or 2nd or 3rd or (1st and 2nd)....

easier way is ---->>>     1 -  (prob of not selected in any company)

Prob of not get selected in any company is 11/12*14/15*9/10 = 77/100

So, prob of getting selected in atleast one company is 1-(77/100)

= 0.23