Question

Mr. A is called to the interviews. There are candidates at the first interview, $ 4 $ at the second and $ 6 $ at the third. If the selection of each candidate is equally likely then the probability that A will be selected for at least one post is?

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
Let us assume that the three candidates are A, B, C respectively to $ 5 $ candidates at the first interview.
In the second interview, there are four and in the third interview, there are six candidates.
Since we need to find the probability of A will be selected for at least one post in that three posts were held.

Formula used: $ P(A \cup B \cup C) = 1 - P(\overline A \cap \overline B \cap \overline C ) $ where $ P(A \cup B \cup C) $ is the overall probability to find at least one post for the candidate one.

Complete step by step answer:
Let the probability that the first interview has been held in $ P(A) = \dfrac{1}{5} $ (there are five candidates for the first interview so the probability is one by five)
Now the probability for a second interview has been held is \[P(B) = \dfrac{1}{4}\](similarly there are four candidates and the probability is one by four).
For the third candidate’s probability is \[P(C) = \dfrac{1}{6}\]and now we are going to substitute the values into the formula of at least one post that is $ P(A \cup B \cup C) = 1 - P(\overline A \cap \overline B \cap \overline C ) $ where the bar is the complement of the given set.
Since the formula can be rewritten as $ P(A \cup B \cup C) = 1 - (P(\overline A ).P(\overline B ).P(\overline C )) $ intersection can be given into the sets of the interview. Since \[P(\overline A ) = 1 - P(A)\]and for all interviews too.
Thus, applying this in the formula we get $ P(A \cup B \cup C) = 1 - [1 - P(A).1 - P(B).1 - P(C)] $ .
Hence applying the given value, we get $ P(A \cup B \cup C) = 1 - [(1 - \dfrac{1}{5})(1 - \dfrac{1}{4})(1 - \dfrac{1}{6})] $ .
Further solving this by cross multiplication we get $ 1 - \dfrac{4}{5}\dfrac{3}{4}\dfrac{5}{6} = \dfrac{1}{2} $ .
Therefore, the probability that A will be selected for at least one post is $ \dfrac{1}{2} $.

Note: we are also able to solve this problem without converting it into values\[P(\overline A ) = 1 - P(A)\].
 $ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) $ where an intersection B is the probability that the first and second interview applicable is likely the same also the probability of A and B and C intersection is the remember three independent events.