AP EAMCET · Maths · Probability
If the probability distribution of a random variable X is as follows, then the mean of X is
| \(X = x_i\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|
| \(\mathrm{P}\left(\mathrm{X} = \mathrm{x}_i\right)\) | \(\mathrm{k}^3\) | \(2\mathrm{k}^3 + \mathrm{k}\) | \(4\mathrm{k} - 10\mathrm{k}^2\) | \(4\mathrm{k} - 1\) |
- A \(\frac{193}{27}\)
- B \(\frac{25}{27}\)
- C \(\frac{23}{27}\)
- D \(\frac{83}{27}\)
Answer & Solution
Correct Answer
(C) \(\frac{23}{27}\)
Step-by-step Solution
Detailed explanation
\(\sum \mathrm{P}(\mathrm{X}=\mathrm{x}_i) = 1 \Rightarrow \mathrm{k}^3 + (2\mathrm{k}^3 + \mathrm{k}) + (4\mathrm{k} - 10\mathrm{k}^2) + (4\mathrm{k} - 1) = 1 \Rightarrow 3\mathrm{k}^3 - 10\mathrm{k}^2 + 9\mathrm{k} - 2 = 0\) For \(k = \frac{1}{3}\):…
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