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MHT CET 2021 20th September Morning Shift

MCQ+2 / -02021

A random variable X has the following probability distribution



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\(\mathrm{X=x}\) 0 1 2 3 4 5 6 7
\(\mathrm{P[X=x]}\) 0 \(\mathrm{k}\) \(\mathrm{2k}\) \(\mathrm{2k}\) \(\mathrm{3k}\) \(\mathrm{k^2}\) \(\mathrm{2k^2}\) \(\mathrm{7k^2+k}\)


then F(4) =

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