AP EAMCET · Maths · Probability

Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P (X = r) = K r^{3}$ where $r = 1, 2, 3, 4$ then

A $K = \frac{1}{100} and P (\frac{1}{2} < X < \frac{5}{2} ∣ X > 1) = \frac{8}{97}$
B $K = \frac{1}{99} and P (\frac{1}{2} < X < \frac{5}{2} ∣ X > 1) = \frac{8}{99}$
C $K = \frac{1}{100} and P (\frac{1}{2} < X < \frac{5}{2} ∣ X > 1) = \frac{8}{99}$
D $K = \frac{1}{100} and P (\frac{1}{2} < X < \frac{5}{2} ∣ X > 1) = \frac{10}{99}$

Verified Solution

Answer & Solution

Correct Answer

(C) $K = \frac{1}{100} and P (\frac{1}{2} < X < \frac{5}{2} ∣ X > 1) = \frac{8}{99}$

Step-by-step Solution

Detailed explanation

According to the given condition, we have random variables 11, 22, 33 and 44 As, PX=r=Kr3 where, r=1, 2, 3, 4 Hence, PX=1=K13=K, PX=2=K23=8K, PX=3=K33=27K and PX=4=K43=64K As, we know, here ∑14PX=r=1 (sum of probabilities)…

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.

Same subject

Explore more questions on app

From AP EAMCET

Explore more questions on app

Let X be a random variable which takes values 1,2,3,4 such that P(X=r)=Kr3 where r=1,2,3,4 then

Step-by-step Solution

Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P (X = r) = K r^{3}$ where $r = 1, 2, 3, 4$ then