ProbabilityHard

Question

A random variable $X$ takes values $0,1,2,3$ with probabilities $\frac{2a + 1}{30},\frac{8a - 1}{30},\frac{4a + 1}{30}$, b respectively, where $a,b \in \mathbf{R}$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of X such that $\sigma^{2} + \mu^{2} = 2$.

Then $\frac{a}{b}$ is equal to :

Options

A.30
B.3
C.60
D.12

Solution

$$x$$ 0 1 2 3
$$p(x)$$ $$\frac{2a + 1}{30}$$ $$\frac{8a - 1}{30}$$ $$\frac{4a + 1}{30}$$ $$b$$

$${\sigma^{2} = \sum x_{i}^{2}p\left( x_{i} \right) - \mu^{2} }{\sigma^{2} + \mu^{2} = \sum x_{i}^{2}p\left( x_{i} \right) }{= 0 + 1\left( \frac{8a - 1}{30} \right) + 4\left( \frac{4a + 1}{30} \right) + 9b }{\Rightarrow \frac{24a + 270\text{ }b + 3}{30} = 2 }{24a + 270\text{ }b = 57}$$

$$\begin{array}{r} 8a + 90\text{ }b = 19\#(1) \end{array}$$

Also

$${\sum p(i) = 1 }{\frac{2a + 1}{30} + \frac{8a - 1}{30} + \frac{4a + 1}{30} + b = 1}$$

$$\begin{array}{r} 14a + 30\text{ }b = 29\#(2) \end{array}$$

Solving (1) & (2)

$$a = 2,\ \text{ }b = \frac{1}{30},\frac{a}{b} = 60$$

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