Given below is the probability distribution of discrete r.v. X \begin{array}{|c|c|c|c|c|c|c|} \hline…

Given below is the probability distribution of discrete r.v. X \begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \mathrm{P}[\mathrm{X}=x] & \mathrm{k} & 0 & 2 \mathrm{k} & 5 \mathrm{k} & \mathrm{k} & 3 \mathrm{k} \\ \hline \end{array} Then $\mathrm{P}[\mathrm{X} \geq 4]=$
  1. $\frac{1}{4}$
  2. $\frac{1}{3}$
  3. $\frac{1}{2}$
  4. $\frac{3}{4}$

Solution

Here $k+0+2 k+5 k+k+3 k=1 \Rightarrow k=\frac{1}{12}$ $\therefore P(X \geq 4)=5 k+k+3 k$ $=9 k=\frac{9}{12}=\frac{3}{4}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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