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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 2 · Set 1 · Q36
Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000…
Smoking increases the risk of lung problems.
A study revealed that $\displaystyle 170$ in $\displaystyle 1000$ males who smoke develop lung complications, while $\displaystyle 120$ out of $\displaystyle 1000$ females who smoke develop lung related problems. In a colony, $\displaystyle 50$ people were found to be smokers of which $\displaystyle 30$ are males. A person is selected at random from these $\displaystyle 50$ people and tested for lung related problems. Based on the given information, answer the following questions :
(i)
What is the probability that selected person is a female ?
(ii)
If a male person is selected, what is the probability that he will not be suffering from lung problems?
(iii)
(a)
A person selected at random is detected with lung complications. Find the probability that selected person is a female.
OR
(b)
A person selected at random is not having lung problems, find the probability that the person is a male.
Marking-scheme solution
Total number of smokers in the colony $\displaystyle =50$,
Number of male smokers $\displaystyle =30$, Number of female smokers $\displaystyle =20$
Let $\displaystyle L$ represent the event that the person is suffering from the lung complications.
(i)
$\displaystyle P(F)=\dfrac{20}{50}=\dfrac{2}{5}$
(ii)
$\displaystyle P\left(L^{\prime} \mid M\right)=1-P(L \mid M)=1-\dfrac{170}{1000}=\dfrac{83}{100}$
(iii)
$\displaystyle P(L \mid M)=\dfrac{170}{1000}=\dfrac{17}{100}$ and $\displaystyle P(L \mid F)=\dfrac{120}{1000}=\dfrac{12}{100}$
Using Baye's Theorem, $\displaystyle P(F \mid L)=\dfrac{P(F) \cdot P(L \mid F)}{P(M) \cdot P(L \mid M)+P(F) \cdot P(L \mid F)}$
$\displaystyle =\dfrac{\frac{2}{5} \cdot \frac{12}{100}}{\frac{3}{5} \cdot \frac{17}{100}+\frac{2}{5} \cdot \frac{12}{100}}$
$\displaystyle =\dfrac{24}{75}$ or $\displaystyle \dfrac{8}{25}$
Here $\displaystyle P\left(L^{\prime} \mid M\right)=\dfrac{83}{100}, P\left(L^{\prime} \mid F\right)=\dfrac{88}{100}$
Using Baye's Theorem, $\displaystyle P\left(M \mid L^{\prime}\right)=\dfrac{P(M) \cdot P\left(L^{\prime} \mid M\right)}{P(M) \cdot P\left(L^{\prime} \mid M\right)+P(F) \cdot P\left(L^{\prime} \mid F\right)}$
$\displaystyle =\dfrac{\frac{3}{5} \cdot \frac{83}{100}}{\frac{3}{5} \cdot \frac{83}{100}+\frac{2}{5} \cdot \frac{88}{100}}$
$\displaystyle =\dfrac{249}{425}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.