Question

Mr. X is called for an interview for 3 separate posts. At the first interview, there are 5 candidates; at the second 4 candidates; at the third 6 candidates. If the selection of each candidate is equally likely, find the probability that Mr. X will be selected for (i) at least one post; (ii) at least two posts.

Answer 1

Assume number of candidates for ${{1}^{st}},{{2}^{nd}}$ and ${{3}^{rd}}$ posts as sets A, B and C respectively, For part (i) apply the formula: - $P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-\left[ P\left( A\cap B \right)+P\left( B\cap C \right)+P\left( C\cap A \right) \right]+P\left( A\cap B\cap C \right)$, where P is the probability of the event. For part (ii), to find the required probability use the expression: - $P\left( A\cap B \right)+P\left( B\cap C \right)+P\left( C\cap A \right)+P\left( A\cap B\cap C \right)$. Remember that for 3 independent events A, B and C we have $P\left( A\cap B\cap C \right)=P\left( A \right)\times P\left( B \right)\times P\left( C \right)$. This is applicable for 2 independent evens also, like $P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right)$.

Complete step-by-step solution
Let us assume the number of candidates for first, second, and third posts is denoted assets A, B, and C respectively.
Therefore, n (A) = 5, n (B) = 4, n (C) = 6, where ‘n’ is the number of candidates or we can say elements in the set.
(i) The probability of Mr. X to be selected for at least one post
= $P\left( A\cup B\cup C \right)$
= $P\left( A \right)+P\left( B \right)+P\left( C \right)-\left[ P\left( A\cap B \right)+P\left( B\cap C \right)+P\left( C\cap A \right) \right]+P\left( A\cap B\cap C \right)$
Since, A, B and C are independent events, therefore,

& \Rightarrow P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right) \\\ & \Rightarrow P\left( B\cap C \right)=P\left( B \right)\times P\left( C \right) \\\ & \Rightarrow P\left( C\cap A \right)=P\left( C \right)\times P\left( A \right) \\\ & \Rightarrow P\left( A\cap B\cap C \right)=P\left( A \right)\times P\left( B \right)\times P\left( C \right) \\\ \end{aligned}$$ Substituting all these values in the probability expression, we have, $$P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-\left[ P\left( A \right)\times P\left( B \right)+P\left( B \right)\times P\left( C \right)+P\left( A \right)\times P\left( C \right) \right]+P\left( A \right)\times P\left( B \right)\times P\left( C \right)$$ Now, $$P\left( A \right)=\dfrac{1}{n\left( A \right)}=\dfrac{1}{5}$$ $$\begin{aligned} & \Rightarrow P\left( B \right)=\dfrac{1}{n\left( B \right)}=\dfrac{1}{4} \\\ & \Rightarrow P\left( C \right)=\dfrac{1}{n\left( C \right)}=\dfrac{1}{6} \\\ \end{aligned}$$ These values when substituted in the probability equation gives, $$\begin{aligned} & \Rightarrow P\left( A\cup B\cup C \right)=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{6}-\left[ \dfrac{1}{4}\times \dfrac{1}{5}+\dfrac{1}{5}\times \dfrac{1}{6}+\dfrac{1}{4}\times \dfrac{1}{6} \right]+\dfrac{1}{4}\times \dfrac{1}{5}\times \dfrac{1}{6} \\\ & \Rightarrow P\left( A\cup B\cup C \right)=\dfrac{1}{2} \\\ \end{aligned}$$ (ii) Here, we have to find the probability of Mr. X to be selected for at least two posts. So, Mr. X can be selected for post A and B, post B and C, post C and A or for all the 3 posts A and B and C. Therefore, the required probability is $$\begin{aligned} & =P\left( A\cap B \right)+P\left( B\cap C \right)+P\left( C\cap A \right)+P\left( A\cap B\cap C \right) \\\ & =P\left( A \right).P\left( B \right)+P\left( B \right).P\left( C \right)+P\left( C \right).P\left( A \right)+P\left( A \right).P\left( B \right).P\left( C \right) \\\ & =\dfrac{1}{5}\times \dfrac{1}{4}+\dfrac{1}{4}\times \dfrac{1}{6}+\dfrac{1}{5}\times \dfrac{1}{6}+\dfrac{1}{4}\times \dfrac{1}{5}\times \dfrac{1}{6} \\\ & =\dfrac{2}{15} \\\ \end{aligned}$$ **Note:** One may note that we have applied all the formulas of set theory. One must remember that for the probability of independent events we take the product of the individual probability of the events. If we will not use this theorem then it will be very difficult to calculate the required probability.