Question

A candidate is selected for an interview for three posts. For the first there are 3 candidates, for the second there are 4 and for the third there are 2. What are the chances of his getting at least one post?
A) \[\dfrac{3}{4}\]
B) \[\dfrac{{11}}{{12}}\]
C) \[\dfrac{1}{4}\]
D) \[\dfrac{7}{8}\]

Answer 1

Hint: Here we will first assume the events of being selected for first post, second post and third post as A, B, C and find the probabilities of A, B and C then we will find the probability of not being selected for any of the posts. Then, in order to find the probability of being selected for at least one post we need to subtract the probability of not being selected from total probability.

Complete step-by-step answer:
Let the event of being selected for first post be A
Let the event of being selected for second post be B
Let the event of being selected for third post be C
Now we know that probability is given by:-
\[{\text{probability}} = \dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total outcomes}}}}\]
For event A:-
Number of favorable outcomes are 1
Number of total outcomes are 3
Hence the probability of being selected for first post i.e. P(A) is given by:
\[P(A) = \dfrac{1}{3}\]
Also, the probability of not being selected for first post i.e. \[P(\overline A )\] is given by:-
\[P(\overline A ) = 1 - P\left( A \right)\]
Putting the value we get:-
\[P(\overline A ) = 1 - \dfrac{1}{3}\]
Solving it further we get:-
\[P(\overline A ) = \dfrac{2}{3}\]…………………………….. (1)
For event B:-
Number of favorable outcomes are 1
Number of total outcomes are 4
Hence the probability of being selected for first post i.e. P (B) is given by:
\[P\left( B \right) = \dfrac{1}{4}\]
Also, the probability of not being selected for first post i.e. \[P(\overline B )\] is given by:-
\[P(\overline B ) = 1 - P\left( B \right)\]
Putting the value we get:-
\[P(\overline B ) = 1 - \dfrac{1}{4}\]
Solving it further we get:-
\[P(\overline B ) = \dfrac{3}{4}\]…………………………….. (2)
For event C:-
Number of favorable outcomes are 1
Number of total outcomes are 2
Hence the probability of being selected for first post i.e. P (C) is given by:
\[P\left( C \right) = \dfrac{1}{2}\]
Also, the probability of not being selected for first post i.e. \[P(\overline C )\] is given by:-
\[P(\overline C ) = 1 - P\left( C \right)\]
Putting the value we get:-
\[P(\overline C ) = 1 - \dfrac{1}{2}\]
Solving it further we get:-
\[P\left( {\overline C } \right) = \dfrac{1}{2}\]…………………………….. (3)
Now we will find the probability of not being selected for any of the post which is given by:-
\[ = P\left( {\overline A } \right).P\left( {\overline B } \right).P\left( {\overline C } \right)\]
Putting values from equation 1, 2 and 3 we get:-
\[{\text{Probability of not being selected for any post}} = \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{1}{2}\]
Solving it further we get:-
\[{\text{Probability of not being selected for any post}} = \dfrac{1}{4}\]
Now we know that in order to find the probability of being selected for at least one post we need to subtract the probability of not being selected from total probability.
Hence,
\[{\text{Probability of being selected for at least one post}} = {\text{1}} - {\text{Probability of not being selected for any post}}\]
Putting the value we get:-
\[{\text{Probability of being selected for at least one post}} = {\text{1}} - \dfrac{1}{4}\]
Solving it further we get:-
\[{\text{Probability of being selected for at least one post}} = \dfrac{3}{4}\]

Hence option A is correct.

Note: Students can also proceed by finding the probability of each case like being selected in only A, only B , only C when making pairs of 2 and then making pairs of 3 events but it would be a very tedious task so the above method should be followed to make calculations easy.