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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
The probability distribution of a random variable x is given as under:
$\text{P}(\text{X}=\text{x})=\begin{cases}\text{k}\text{x}^2 & \text{for}\text{ x}= 1,2,3\\2\text{kx} & \text{for}\text{ x } =4,5,6\\0&\text{otherwise} \end{cases}$
where k is a constant. Calculate
  1. $\text{E}(\text{X})$
  2. $\text{E}(3\text{X}^2)$
  3. $\text{P}(\text{X}\geq4)$
✓

Answer

Given condition can be represented as below:
X
1
2
3
4
5
6
Otherwise
P(X)
k
4k
9k
8k
10k
12k
0
We know that, $\sum\text{P}_{\text{i}}=1$ $\Rightarrow44\text{k}=1$ $\Rightarrow\text{k}=\frac{1}{44}$ $\therefore\sum\text{XP}(\text{X})=\text{k}+8\text{k}+27\text{k}+32\text{k}+50\text{k}+72\text{k}+0$ $=190\text{k}=190\times\frac{1}{44}=\frac{95}{22}$
  1. So, $\text{E}(\text{X})=\sum\text{XP}(\text{X})=\frac{95}{22}=4.32$
  2. Also, $\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})=\text{k}+16\text{k}+128\text{k}+250\text{k}+432$
$=908\text{k}=908\times\frac{1}{44}$ $\Big[\because\text{k}=\frac{1}{44}\Big]$
$=20.636=20.64$ (approx)
$\therefore\text{E}(3\text{X}^2)=3\text{E}(\text{X}^2)=3\times20.64$
$=61.92=61.9$
  1. $\text{P}(\text{X}\geq4)=\text{P}(\text{X}=4)+\text{P}(\text{X}=5)+\text{P}(\text{X}=6)$
$=8\text{k}+10\text{k}+12\text{k}=30\text{k}$
$=30\cdot\frac{1}{44}=\frac{15}{22}$

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