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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 3 · Set 1 · Q37
A survey was conducted to find out the success rate of students who qualified the entrance examination by dropping a year after class XII. As per the…
A survey was conducted to find out the success rate of students who qualified the entrance examination by dropping a year after class XII.
As per the data collected, $\displaystyle 40$% students appearing in the examination were dropouts and the remaining students were regular students of class XII. Of the dropouts, $\displaystyle 5$% qualify the examination while $\displaystyle 10$% of the regular students qualify the examination. Based on the above information, answer the following questions.
(i)
Find the probability that a student selected at random is a regular student.
(ii)
A student is selected at random from a group of dropout students. What is the probability that the student will not qualify the examination ?
(iii)
(a)
A student selected at random qualified the examination. Find the probability that student is not a dropout.
OR
(b)
A student selected at random did not qualify the examination. Find the probability that the student was a regular student.
Marking-scheme solution
(i)
Probability that a student selected at random is a regular student $\displaystyle =\dfrac{3}{5}$
(ii)
Probability that the selected student from the group of dropout will not qualify the examination $\displaystyle =\dfrac{95}{100}$ or $\displaystyle \dfrac{19}{20}$
(iii)
Let A : The student has qualified the examination.
$\displaystyle E_{1}$ : The student appearing in the examination is dropout.
$\displaystyle E_{2}$ : The student is regular.
Using Bayes' Theorem, $\displaystyle P\left(\dfrac{E_{1}{}^{\prime}}{A}\right)=P\left(\dfrac{E_{2}}{A}\right)=\dfrac{\frac{60}{100} \times \frac{10}{100}}{\frac{40}{100} \times \frac{5}{100}+\frac{60}{100} \times \frac{10}{100}}$
$\displaystyle =\dfrac{3}{4}$
$\displaystyle P\left(\dfrac{E_{2}}{A^{\prime}}\right)=\dfrac{\frac{60}{100} \times \frac{90}{100}}{\frac{40}{100} \times \frac{95}{100}+\frac{60}{100} \times \frac{90}{100}}$
$\displaystyle =\dfrac{27}{46}$
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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.