Question

A candidate is selected for the 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.

Answer 1

Hint: Probability is a measure of the likelihood that an event will happen.
\[Probability{\text{ }} = \;\dfrac{{favorable\;outcomes}}{{possible\;outcomes}}\]
You cannot answer it without assuming very unrealistic things (which are not mentioned). Independence between interviews of any of the people

Complete step-by-step answer:
P(at least one offer) = 1 - p(no offer) = 1 - p(no offer for post 1) Xp(no offer for post 2) Xp(no offer for post 3) by assuming 3 interviews are independent and each person has equal opportunity for each interview.
The total number of elementary events of a trial are known as the total number of cases.
Desired outcome of an elementary event is called a favourable event.
For the first company there are 12 candidates (given)
Probability of being not selected in first company = \[\dfrac{{Favourable\;candidate}}{{total\;no\;of\;candidate}} = \dfrac{{11}}{{12}}..................(1)\]
For the second company there are 15 candidates (given)
Probability of being not selected in second company = \[\dfrac{{Favourable\;candidate}}{{total\;no\;of\;candidate}} = \dfrac{{14}}{{15}}..................(1)\]
For the third company there are 10 candidates (given)
Probability of being not selected in third company = \[\dfrac{{Favourable\;candidate}}{{total\;no\;of\;candidate}} = \dfrac{9}{{10}}..................(1)\]
SO , overall probability not being selected with any of the company = p(no offer for post 1) X p(no offer for post 2) X p(no offer for post 3)
By substituting the value from equation (1), (2) and (3)
\[Overall{\text{ }}probability{\text{ }}not{\text{ }}being{\text{ }}selected{\text{ }}with{\text{ }}any{\text{ }}of{\text{ }}the{\text{ }}company{\text{ }} = \dfrac{{11}}{{12}} \times \dfrac{{14}}{{15}} \times \dfrac{9}{{10}} = \dfrac{{231}}{{300}}\]
P(at least one offer) = 1 - p(no offer) = 1 - p(no offer for post 1) Xp(no offer for post 2) Xp(no offer for post 3) by assuming 3 interviews are independent and each person has equal opportunity for each interview.
By substituting the value from equation above, we get:
\[Overall{\text{ }}probability{\text{ }}with{\text{ }}atleast{\text{ }}one{\text{ }}offer{\text{ }} = 1 - \dfrac{{231}}{{300}} = \dfrac{{69}}{{300}} = \dfrac{{23}}{{100}}\]

Therefore, the probability that he is selected in at least one of the companies is \[\dfrac{{23}}{{100}}\]

Note: If all outcomes are favorable for a certain event, its probability is 1. For example, the probability of rolling a 6 or lower on one die is = 1.
If none of the possible outcomes are favorable for a certain event (a favorable outcome is impossible), the probability is 0. For example, the probability of rolling a 7 on one die is = 0.