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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 5 · Set 1 · Q31
The random variable X has the following probability distribution where a and b are some constants: If the mean E ( X )=3, then find values of a and b…
The random variable X has the following probability distribution where a and $\displaystyle b$ are some constants :
If the mean $\displaystyle \mathrm{E}(\mathrm{X})=3$, then find values of a and b and hence determine $\displaystyle \mathrm{P}(\mathrm{X} \geq 3)$.
| $\displaystyle \mathbf{X}$ | $\displaystyle 1$ | $\displaystyle 2$ | $\displaystyle 3$ | $\displaystyle 4$ | $\displaystyle 5$ |
| $\displaystyle \mathbf{P}(\mathbf{X})$ | $\displaystyle 0.2$ | a | a | $\displaystyle 0.2$ | b |
Marking-scheme solution
$$\begin{aligned}
& \mathrm{E}(\mathrm{X})=0.2+2 \mathrm{a}+3 \mathrm{a}+0.8+5 b=5 \mathrm{a}+5 b+1 \\
& \sum \mathrm{p}_{\mathrm{i}}=1 \Rightarrow 2 \mathrm{a}+b=0.6 \\
& \mathrm{E}(\mathrm{X})=3 \Rightarrow 5 \mathrm{a}+5 b=2 \text { and, } \\
& \text { solving the two equations, we get, } \mathrm{a}=\frac{1}{5}, \mathrm{~b}=\frac{1}{5} \\
& \mathrm{P}(\mathrm{X} \geq 3)=1-[\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)]=1-[0.2+\mathrm{a}]=1-\frac{2}{5}=\frac{3}{5}
\end{aligned}
$$
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CBSE Class 12 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.