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MHT CET · Maths · Probability

A random variable X has the following probability distribution
A random variable X has the following probability distribution
\(\begin{array}{|r|c|c|c|c|c|}
\hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}): & \mathrm{k} & 2 \mathrm{k} & 4 \mathrm{k} & 2 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}\)
then the value of \(\mathrm{P}(1 \leqslant \mathrm{X}<4 / \mathrm{X} \leqslant 2)=\)
then the value of \(\mathrm{P}(1 < \mathrm{X} < 4 / \mathrm{X} < 2)=\)

  1. A \(\frac{5}{6}\)
  2. B \(\frac{6}{7}\)
  3. C \(\frac{7}{8}\)
  4. D \(\frac{8}{9}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{6}{7}\)

Step-by-step Solution

Detailed explanation

\(\sum \mathrm{P}(\mathrm{X}) = 1 \Rightarrow \mathrm{k} + 2\mathrm{k} + 4\mathrm{k} + 2\mathrm{k} + \mathrm{k} = 1 \Rightarrow 10\mathrm{k} = 1 \Rightarrow \mathrm{k} = \frac{1}{10}\) \(\mathrm{P}(\mathrm{X}=0) = \frac{1}{10}, \mathrm{P}(\mathrm{X}=1) = \frac{2}{10}, \mathrm{P}(\mathrm{X}=2) = \frac{4}{10}, \mathrm{P}(\mathrm{X}=3) = \frac{2}{10}, \mathrm{P}(\mathrm{X}=4) = \frac{1}{10}\)