Skip to content
Question of 165

Q.The range of a random variable xx is {0,1,2}\{0, 1, 2\}. Given that P(x=0)=3c3P(x = 0) = 3c^3, P(x=1)=4c−10c2P(x = 1) = 4c - 10c^2, P(x=2)=5c−1P(x = 2) = 5c - 1. (i)(i) Find the value of cc. (ii)(ii) P(x<1)P(x < 1), P(1<x≤2)P(1 < x \le 2) and P(0<x≤3)P(0 < x \le 3).

Telangana TsbieTelangana Board of Intermediate Education 2023Subjective· 7mImportance★★★★★
0% · 0/165 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Summation gives c=13c=\tfrac13 (the only valid root); then P(x<1)=19P(x<1)=\tfrac19, P(1<x≤2)=23P(1<x\le2)=\tfrac23, P(0<x≤3)=89P(0<x\le3)=\tfrac89.

Since ∑P=1\sum P = 1:

3c3+(4c−10c2)+(5c−1)=13c^3 + (4c - 10c^2) + (5c - 1) = 1

3c3−10c2+9c−2=03c^3 - 10c^2 + 9c - 2 = 0.

Factor: c=1c = 1 is a root, giving (c−1)(3c2−7c+2)=0(c-1)(3c^2 - 7c + 2) = 0, and 3c2−7c+2=0⇒c=23c^2 - 7c + 2 = 0 \Rightarrow c = 2 or c=13c = \tfrac13.

Probabilities must lie in [0,1][0,1]. For c=1c=1, P(x=0)=3P(x=0)=3; for c=2c=2, P(x=0)=24P(x=0)=24 — both invalid. Only c=13c = \dfrac13 works:

P(x=0)=3(13)3=19,P(x=1)=43−109=29,P(x=2)=53−1=23P(x=0) = 3\left(\tfrac13\right)^3 = \dfrac19,\quad P(x=1) = \dfrac43 - \dfrac{10}{9} = \dfrac29,\quad P(x=2) = \dfrac53 - 1 = \dfrac23.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.