Mht Cet
Probability
MHT CET 2025 20th April Morning Shift
MCQ+2 / -02025
A random variable $X$ has the following probability distribution
$$ \begin{array}{|l|c|c|c|c|c|} \hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}): & \mathrm{k} & 2 \mathrm{k} & 4 \mathrm{k} & 2 \mathrm{k} & \mathrm{k} \\ \hline \end{array} $$
then the value of $\mathrm{P}(1 \leqslant \mathrm{X}<4 \mid \mathrm{X} \leqslant 2)=$
