MHT CET · Maths · Probability
If a random variable \(X\) has the following probability distribution of \(X\)
\(\begin{array}{|l|c|c|c|c|c|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0 & \mathrm{k} & 2 \mathrm{k} & 2 \mathrm{k} & 3 \mathrm{k} & \mathrm{k}^2 & 2 \mathrm{k}^2 & 7 \mathrm{k}^2+\mathrm{k} \\ \hline \end{array}\)
Then \(\mathrm{P}(\mathrm{X} \geqslant 6)=\)
- A \(\frac{19}{100}\)
- B \(\frac{81}{100}\)
- C \(\frac{9}{100}\)
- D \(\frac{91}{100}\)
Answer & Solution
Correct Answer
(A) \(\frac{19}{100}\)
Step-by-step Solution
Detailed explanation
\( \sum P(X=x) = 1 \) \( k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1 \)
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