Question

Smoking increases the risk of lung problems.

A study revealed that 170 in 1000 males who smoke develop lung related complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males.

A person is selected at random from these 50 people and tested for lung related problems.

Based on the given information, answer the following questions:

1. What is the probability that selected person is a female? 1
2. If a male person is selected, what is the probability that he will not be suffering from lung problems? 1
3. 1. A person selected at random is detected with lung complications. Find the probability that selected person is a female. 2
      OR
   2. A person selected at random is not having lung problems, find the probability that the person is a male. 2

Answer 1

(i) Given, Total no. of people be 50

no. of males = 30

no. of females = 20

P(F) = `"No. of favourable outcome"/"No. of toal outcome"` 

`20/50`

= 0.4

(ii) Let C be the event that person suffers from lung complications.

C¹ be the event that person not suffers from lung complications.

P(c/m) = `(170)/(1000)=0.17`;

`P(C^1/m)=1-P(C/m)`

= 1 – 0.17

= 0.83

(iii) (a) 

P(F) = 0.4; P(m) = `30/50` = 0.6

`P(C/F) = 120/1000=0.120`; 

`P(C^1/F)=1-0.120`

= 0.88

 By using total probability,

P(C) = `P(M)*P(C/M)+P(F)*P(C/F)`

= 0.6 × 0.17 + 0.4 × 0.120

= 0.102 + 0.048

= 0.15 

`P(F/C)=P(F)*(P(C/F))/(P(C))`

= `(0.4xx0.120)/0.15`

= `0.048/0.15`

= 0.32

(iii) (b) 

 `P(C^1)=P(M)*P(M/C^1)+P(F)*P(F/C^1)`   ...{By using total probability}

= 0.6 × 0.83 + 0.4

= 0.83

`P(M/C^1)=(0.6xx0.83)/0.85`

= `498/850`

= 0.585