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Hint- In this question we have to find the probability of a smoker, if a male is first selected. So, we will be using the formula for conditional probability, that is $P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$. This property will help us simplify things up and will eventually help us reach the answer.

Complete step-by-step answer:

In the question, we have been given that, the probability of selecting a male or smoker is $\dfrac{7}{{10}}$, a male smoker is $\dfrac{2}{5}$ and a male, if a smoker is already selected, is $\dfrac{2}{3}$.

So, let the event of selecting a male be A and the event of selecting a smoker be B.

Then, we will have

The probability of selecting a male or smoker = $P\left( {A \cup B} \right)$ = $\dfrac{7}{{10}}$………. Equation (1)

The probability of selecting a male smoker = $P\left( {A \cap B} \right)$ = $\dfrac{2}{5}$…………….. Equation (2)

The probability of selecting a male, if a smoker is already selected = $P\left( {\frac{A}{B}} \right)$ = $\dfrac{2}{3}$……….. Equation (

And we have to find the probability of selecting a smoker, if a male is first selected = $P\left( {\dfrac{B}{A}} \right)$.

As we know that formula for conditional probability is: $P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$

So, $P\left( B \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( {\dfrac{A}{B}} \right)}}$

$ \Rightarrow P\left( B \right) = \dfrac{2}{5} \times \dfrac{3}{2}$

$ \Rightarrow P\left( B \right) = \dfrac{3}{5}$

Now, using sets we know that $P\left( A \right) = P\left( {A \cup B} \right) + P\left( {A \cap B} \right) - P\left( B \right)$

So, $P\left( A \right) = \dfrac{7}{{10}} + \dfrac{2}{5} - \dfrac{3}{5}$

$ \Rightarrow P\left( A \right) = \dfrac{{7 + 4 - 6}}{{10}} = \dfrac{1}{2}$

Now, we will find the probability of selecting a smoker, if a male is first selected = $P\left( {\dfrac{B}{A}} \right)$.

Using conditional probability, we get

$P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$

$ \Rightarrow P\left( {\dfrac{B}{A}} \right) = \dfrac{2}{5} \times \dfrac{2}{1} = \dfrac{4}{5}$

Hence, the option C is correct.

Note- Whenever we face such types of problems the key point to remember is that we need to have a good grasp over probability, conditional probability and related concepts. The conditional probability has been discussed above. This formula helps in the simplification and getting on the right track to reach the answer.

Complete step-by-step answer:

In the question, we have been given that, the probability of selecting a male or smoker is $\dfrac{7}{{10}}$, a male smoker is $\dfrac{2}{5}$ and a male, if a smoker is already selected, is $\dfrac{2}{3}$.

So, let the event of selecting a male be A and the event of selecting a smoker be B.

Then, we will have

The probability of selecting a male or smoker = $P\left( {A \cup B} \right)$ = $\dfrac{7}{{10}}$………. Equation (1)

The probability of selecting a male smoker = $P\left( {A \cap B} \right)$ = $\dfrac{2}{5}$…………….. Equation (2)

The probability of selecting a male, if a smoker is already selected = $P\left( {\frac{A}{B}} \right)$ = $\dfrac{2}{3}$……….. Equation (

And we have to find the probability of selecting a smoker, if a male is first selected = $P\left( {\dfrac{B}{A}} \right)$.

As we know that formula for conditional probability is: $P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$

So, $P\left( B \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( {\dfrac{A}{B}} \right)}}$

$ \Rightarrow P\left( B \right) = \dfrac{2}{5} \times \dfrac{3}{2}$

$ \Rightarrow P\left( B \right) = \dfrac{3}{5}$

Now, using sets we know that $P\left( A \right) = P\left( {A \cup B} \right) + P\left( {A \cap B} \right) - P\left( B \right)$

So, $P\left( A \right) = \dfrac{7}{{10}} + \dfrac{2}{5} - \dfrac{3}{5}$

$ \Rightarrow P\left( A \right) = \dfrac{{7 + 4 - 6}}{{10}} = \dfrac{1}{2}$

Now, we will find the probability of selecting a smoker, if a male is first selected = $P\left( {\dfrac{B}{A}} \right)$.

Using conditional probability, we get

$P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$

$ \Rightarrow P\left( {\dfrac{B}{A}} \right) = \dfrac{2}{5} \times \dfrac{2}{1} = \dfrac{4}{5}$

Hence, the option C is correct.

Note- Whenever we face such types of problems the key point to remember is that we need to have a good grasp over probability, conditional probability and related concepts. The conditional probability has been discussed above. This formula helps in the simplification and getting on the right track to reach the answer.

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