TS EAMCET · Maths · Probability

A random variable $X$ takes the values $1, 2, 3$ and $4$ such that $2 P (X = 1) = 3 P (X = 2) = P (X = 3) = 5 P (X = 4)$ . If $σ^{2}$ is the variance and $μ$ is the mean of $X$ , then $σ^{2} + μ^{2} =$

A $\frac{421}{61}$
B $\frac{570}{61}$
C $\frac{149}{61}$
D $\frac{3480}{3721}$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{421}{61}$

Step-by-step Solution

Detailed explanation

2PX=1=3PX=2=PX=3=5PX=4 Let PX=3=α. ∑PX=1 ⇒P1+P2+P3+P4=1 ⇒α2+α3+α+α5=1 ⇒6130α=1 ⇒α=3061 Similarly, all values can be found out. x 1 2 3 4 PX 1561 1061 3061 661 Mean μ=EX=∑X=14XPXX =11561+21061+33061+4661…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A random variable X takes the values 1, 2, 3 and 4 such that 2PX=1=3PX=2=PX=3=5PX=4. If σ2 is the variance and μ is the mean of X, then σ2+μ2=

Step-by-step Solution

A random variable $X$ takes the values $1, 2, 3$ and $4$ such that $2 P (X = 1) = 3 P (X = 2) = P (X = 3) = 5 P (X = 4)$ . If $σ^{2}$ is the variance and $μ$ is the mean of $X$ , then $σ^{2} + μ^{2} =$