TS EAMCET · Maths · Probability
A random variable \(\mathrm{X}\) has the range \(\{0,1,2, \ldots\}\). If \(\mathrm{P}(X=r)=k(1+r) 3^{-r}\), for \(r=0,1,2, \ldots\). where \(k>0\) is a real number, then \(P(X=0)+P(X=1)+P(X=2)=\)
- A \(\frac{4}{9}\)
- B \(\frac{8}{9}\)
- C \(\frac{2}{3}\)
- D -
Answer & Solution
Correct Answer
(B) \(\frac{8}{9}\)
Step-by-step Solution
Detailed explanation
Given 9 robability function \(\mathrm{P}(\mathrm{X}=r)=\mathrm{K}(1+r) 3^{-r}\). The sum of all probabilities is equal to 1 . So, \(\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\) \(=1\)…
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