Verified Solution • 2025–26 Session moderate

QJEEM21Ma3bf9: JEE Main 2021 Mathematics - Class 12 JEEM21Ma3bf9 Solution: In a group of 400 people, 160 are smokers and non-vegetarian;…

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 :

Multiple Choice Options

  • A 1445{{14} \over {45}}
  • B 845{{8} \over {45}}
  • C 745{{7} \over {45}}
  • D 2845{{28} \over {45}}

Step-by-Step Verified Solution

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)=160400P(A) = {{160} \over {400}}

P(B)=100400P(B) = {{100} \over {400}}

P(C)=140400P(C) = {{140} \over {400}}

P(EA)=35100P\left( {{E \over A}} \right) = {{35} \over {100}}

P(EB)=20100P\left( {{E \over B}} \right) = {{20} \over {100}}

P(EC)=10100P\left( {{E \over C}} \right) = {{10} \over {100}}

To find

P(AE)=P(A)P(EA)P(A).P(EA)+P(B).P(EB)+P(C).P(EC)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)}}

=160400×35100160400×35100×100400×20200+140400×10100= {{{{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}}}}

=2845= {{28} \over {45}}

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