If the p.m.f. is given by $\mathrm{P}(\mathrm{X})=\mathrm{k}\left(\begin{array}{c}4 \\ x\end{array}\right)$,…

If the p.m.f. is given by $\mathrm{P}(\mathrm{X})=\mathrm{k}\left(\begin{array}{c}4 \\ x\end{array}\right)$, for $x=0,1,2,3,4, \mathrm{k}>0$ $=0$, otherwise then the value of $\mathrm{k}$ is
  1. $\frac{3}{16}$
  2. $\frac{7}{16}$
  3. $\frac{1}{16}$
  4. $\frac{5}{16}$

Solution

For $P(X=0)=k\left(\begin{array}{l}4 \\ 0\end{array}\right)=k \times{ }^{4} C_{0}=k \times 1=k$ $\therefore P(X=1)=k\left(\begin{array}{l}4 \\ 1\end{array}\right)=k \times{ }^{4} C_{1}=k(4)=4 k$ $\begin{array}{l} P(X=2)=k\left(\begin{array}{l} 4 \\ 2 \end{array}\right)=k \times{ }^{4} C_{2}=k(6)=6 k \\ P(X=3)=k\left(\begin{array}{l} 4 \\ 3 \end{array}\right)=k x^{4} C_{3}=k(4)=4 k \\ P(x=4)=k\left(\begin{array}{l} 4 \\ 4 \end{array}\right)=k \times 4 C_{4}=(k)(1)=k \end{array}$ Since $P(X)$ is p.m.f. $k+4 k+6 k+4 k+k=1 \Rightarrow 16 k=1 \Rightarrow k=\frac{1}{16}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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