AP EAMCET · Maths · Probability
The range of a random variable \(X\) is \(\{0,1,2\}\). If \(P(X=0)\) \(=3 \mathrm{C}^3, \mathrm{P}(\mathrm{X}=1)=4 \mathrm{C}-10 \mathrm{C}^2\) and \(\mathrm{P}(\mathrm{X}=2)=5 \mathrm{C}-1\), then the value of \(\mathrm{C}\) is
- A \(\frac{2}{3}\)
- B \(\frac{1}{3}\)
- C . \(\frac{5}{3}\)
- D \(\frac{4}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { } \because P(X=0)+P(X=1)+P(X=2)=1 \\ & \Rightarrow 3 c^3+4 c-10 c^2+5 c-1=1 \\ & \Rightarrow 3 c^3+4 c-10 c^2+5 c-2=0 \\ & \Rightarrow(c-1)(3 c-1)(c-2)=0 \\ & \Rightarrow c=\frac{1}{3}, 1,2 . \end{aligned} \] When \(c=2\), Then \(P(X=2)=10-1=9>1\)…
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