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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
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$=0$, otherwise then the value of $\mathrm{k}$ is
Solution:
1908 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{16}$
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.
$$
k+4 k+6 k+4 k+k=1 \Rightarrow 16 k=1 \Rightarrow k=\frac{1}{16}
$$
$\therefore P(X=1)=k\left(\begin{array}{l}4 \\ 1\end{array}\right)=k \times{ }^{4} C_{1}=k(4)=4 k$
$$
\begin{array}{l}
P(X=2)=k\left(\begin{array}{l}
4 \\
2
\end{array}\right)=k \times{ }^{4} C_{2}=k(6)=6 k \\
P(X=3)=k\left(\begin{array}{l}
4 \\
3
\end{array}\right)=k x^{4} C_{3}=k(4)=4 k \\
P(x=4)=k\left(\begin{array}{l}
4 \\
4
\end{array}\right)=k \times 4 C_{4}=(k)(1)=k
\end{array}
$$
Since $P(X)$ is p.m.f.
$$
k+4 k+6 k+4 k+k=1 \Rightarrow 16 k=1 \Rightarrow k=\frac{1}{16}
$$
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