Question 9 Exercise 6.5

Solutions of Question 9 of Exercise 6.5 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

A'jmal and Bushra appear in an interview for $2$ vacancies. The probability of their selection being $\dfrac{1}{7}$ and $\dfrac{1}{5}$ respectively. Find the probability that both will be selected.

\begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text { Bushra selection })&=\dfrac{1}{5} \\ \Rightarrow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align} Both are selected

Since the selection of one does not depend on the other sclection,

therefore these two events are independent. Thus \begin{align}P(\text { Both are selected })&=P(A) \times P(B) \\ \Rightarrow P(\text { Both are selected }) & =\dfrac{1}{7} \times \dfrac{1}{5}=\dfrac{1}{35}\end{align}

A'jmal and Bushra appear in an interview for $2$ vacancies. The probability of their selection being $\dfrac{1}{7}$ and $\dfrac{1}{5}$ respectively. Find the probability that Only one is selected.

\begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text { Bushra selection })&=\dfrac{1}{5} \\ \Rightarrow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align} Only one is selected

In this case there are two chance.

it may be Ajmal selected Bushra not selected that is $P(A \cap \bar{B})$.

The second option is it may be Bushra selected but Ajmal not that is $P(\bar{A} \cup B)$.

Also these both are independent events.

Thus the probability that only one is \begin{align} \text { selected is: }&=P(A \cap \bar{B})+P(\bar{A} \cup B) \\ & =P(A) \times P(\bar{B})+P(\bar{A}) \times P(B) \\ & =\dfrac{1}{7} \cdot \dfrac{4}{5}+\dfrac{6}{7} \cdot \dfrac{1}{5} \\ & =\dfrac{10}{35}=\dfrac{2}{7} \end{align}

A'jmal and Bushra appear in an interview for $2$ vacancies. The probability of their selection being $\dfrac{1}{7}$ and $\dfrac{1}{5}$ respectively. Find the probability that none will be sclected.

\begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text { Bushra selection })&=\dfrac{1}{5} \\ \Rightarrow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align} None will be selected.

Here we have to find probability that none of both is selected that is: \begin{align}P(\bar{A} \cap \bar{B})&=P(\bar{A}) \times P(\bar{B}) \\ \Rightarrow P(\bar{A} \cap \bar{B})&=\dfrac{6}{7} \times \dfrac{4}{5}=\dfrac{24}{35}\end{align}

A'jmal and Bushra appear in an interview for $2$ vacancies. The probability of their selection being $\dfrac{1}{7}$ and $\dfrac{1}{5}$ respectively. Find the probability that at least one of them will be selected.

\begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text { Bushra selection })&=\dfrac{1}{5} \\ \Rightarrow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align} At least one of them will be selected.

If at least one of them is selected, it means either Ajmal is selected,

Bushra is selected or both of them to be selected.

Thus the probability of selecting at least one of them is: \begin{align}P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cup B)&=\dfrac{1}{7}+\dfrac{1}{5}-\dfrac{1}{7} \times \dfrac{1}{5} \\ \because \quad P(A \cap B)&=P(A) \times P(B)\\ \Rightarrow P(A \cup B)&=\dfrac{5+7-1}{35}=\dfrac{11}{35} \end{align}

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