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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 1 · Q30
(a) The probability distribution for the number of students being absent in a class on a Saturday is as follows: Where X is the number of students…
The probability distribution for the number of students being absent in a class on a Saturday is as follows :
Where X is the number of students absent.
| $\displaystyle \mathbf{X}$ | $\displaystyle 0$ | $\displaystyle 2$ | $\displaystyle 4$ | $\displaystyle 5$ |
| $\displaystyle \mathbf{P}(\mathbf{X})$ | p | $\displaystyle 2$ p | $\displaystyle 3$ p | p |
(i)
Calculate p.
(ii)
Calculate the mean of the number of absent students on Saturday.
For the vacancy advertised in the newspaper, $\displaystyle 3000$ candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is $\displaystyle 0.4$ and that a female getting a distinction is $\displaystyle 0.35$ . Find the probability that the candidate chosen at random will have a distinction in the written test.
Marking-scheme solution
\[\begin{aligned}
& \text { (i)Since } \sum P(X)=1 \Rightarrow p+2 p+3 p+p=1 \\
& \Rightarrow p=\frac{1}{7} \\
& \begin{array}{l}
\text { (ii) Mean }=\sum X . P(X)=0(p)+2(2 p)+4(3 p)+5(p) \\
=21 p=21\left(\frac{1}{7}\right)=3
\end{array}
\end{aligned}
\]
Let $\displaystyle \mathrm{E}_{1}$ : Theapplicant is amale
$\displaystyle \mathrm{E}_{2}$ : Theapplicant is a female
A:The candidate chosen will have distinction in the written test.
\[\begin{aligned}
P\left(\mathrm{E}_{1}\right) & =\frac{1}{3}, P\left(\mathrm{E}_{2}\right)=\frac{2}{3}, P\left(A \mid \mathrm{E}_{1}\right)=0.4, P\left(A \mid \mathrm{E}_{2}\right)=0.35 \\
\therefore P(A) & =P\left(\mathrm{E}_{1}\right) P\left(A \mid \mathrm{E}_{1}\right)+P\left(\mathrm{E}_{2}\right) P\left(A \mid \mathrm{E}_{2}\right) \\
& =\frac{1}{3} \times 0.4+\frac{2}{3} \times 0.35 \\
& =\frac{11}{30}
\end{aligned}
\]
\[\begin{aligned}
P\left(\mathrm{E}_{1}\right) & =\frac{1}{3}, P\left(\mathrm{E}_{2}\right)=\frac{2}{3}, P\left(A \mid \mathrm{E}_{1}\right)=0.4, P\left(A \mid \mathrm{E}_{2}\right)=0.35 \\
\therefore P(A) & =P\left(\mathrm{E}_{1}\right) P\left(A \mid \mathrm{E}_{1}\right)+P\left(\mathrm{E}_{2}\right) P\left(A \mid \mathrm{E}_{2}\right) \\
& =\frac{1}{3} \times 0.4+\frac{2}{3} \times 0.35 \\
& =\frac{11}{30}
\end{aligned}
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.