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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 1 · Set 1 · Q30
(a) The probability distribution of a random variable X is given below: (i) Find the value of k. (ii) Find P (1 ≤ X <3). (iii) Find E ( X ), the mean…
The probability distribution of a random variable X is given below :
| X | $\displaystyle 1$ | $\displaystyle 2$ | $\displaystyle 3$ |
| $\displaystyle \mathrm{P}(\mathrm{X})$ | $\displaystyle \frac{\mathrm{k}}{2}$ | $\displaystyle \frac{\mathrm{k}}{3}$ | $\displaystyle \frac{\mathrm{k}}{6}$ |
(i)
Find the value of k .
(ii)
Find $\displaystyle \mathrm{P}(1 \leq \mathrm{X}<3)$.
(iii)
Find $\displaystyle \mathrm{E}(\mathrm{X})$, the mean of X .
A and B are independent events such that $\displaystyle \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=\frac{1}{4}$ and $\displaystyle \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\frac{1}{6}$. Find $\displaystyle \mathrm{P}(\mathrm{A})$ and $\displaystyle \mathrm{P}(\mathrm{B})$.
Marking-scheme solution
(a)(i)
$\displaystyle \frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{3}+\frac{\mathrm{k}}{6}=1$
Gives $\displaystyle \mathrm{k}=1$
(ii)
$\displaystyle \mathrm{P}(1 \leq \mathrm{X}<3)=\frac{5 \mathrm{k}}{6}=\frac{5}{6}$
(iii)
$\displaystyle \mathrm{E}(\mathrm{X})=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=\frac{\mathrm{k}}{2}+\frac{2 \mathrm{k}}{3}+\frac{\mathrm{k}}{2}=\frac{5 \mathrm{k}}{3}$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.