JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 13)
In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is :
$${{14} \over {45}}$$
$${{8} \over {45}}$$
$${{7} \over {45}}$$
$${{28} \over {45}}$$
Explanation
Consider following events
A : Person chosen is a smoker and non vegetarian.
B : Person chosen in a smoker and vegetarian.
C : Person chosen is a non-smoker and vegetarian.
E : Person chosen has a chest disorder
Given
$$P(A) = {{160} \over {400}}$$
$$P(B) = {{100} \over {400}}$$
$$P(C) = {{140} \over {400}}$$
$$P\left( {{E \over A}} \right) = {{35} \over {100}}$$
$$P\left( {{E \over B}} \right) = {{20} \over {100}}$$
$$P\left( {{E \over C}} \right) = {{10} \over {100}}$$
To find
$$P\left( {{A \over E}} \right) = {{P(A)P\left( {{E \over A}} \right)} \over {P(A).P\left( {{E \over A}} \right) + P(B).P\left( {{E \over B}} \right) + P(C).P\left( {{E \over C}} \right)}}$$
$$ = {{{{160} \over {400}} \times {{35} \over {100}}} \over {{{160} \over {400}} \times {{35} \over {100}} \times {{100} \over {400}} \times {{20} \over {200}} + {{140} \over {400}} \times {{10} \over {100}}}}$$
$$ = {{28} \over {45}}$$
A : Person chosen is a smoker and non vegetarian.
B : Person chosen in a smoker and vegetarian.
C : Person chosen is a non-smoker and vegetarian.
E : Person chosen has a chest disorder
Given
$$P(A) = {{160} \over {400}}$$
$$P(B) = {{100} \over {400}}$$
$$P(C) = {{140} \over {400}}$$
$$P\left( {{E \over A}} \right) = {{35} \over {100}}$$
$$P\left( {{E \over B}} \right) = {{20} \over {100}}$$
$$P\left( {{E \over C}} \right) = {{10} \over {100}}$$
To find
$$P\left( {{A \over E}} \right) = {{P(A)P\left( {{E \over A}} \right)} \over {P(A).P\left( {{E \over A}} \right) + P(B).P\left( {{E \over B}} \right) + P(C).P\left( {{E \over C}} \right)}}$$
$$ = {{{{160} \over {400}} \times {{35} \over {100}}} \over {{{160} \over {400}} \times {{35} \over {100}} \times {{100} \over {400}} \times {{20} \over {200}} + {{140} \over {400}} \times {{10} \over {100}}}}$$
$$ = {{28} \over {45}}$$
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