MHT CET · Maths · Probability
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
- A \(\frac{3}{16}\)
- B \(\frac{7}{16}\)
- C \(\frac{1}{16}\)
- D \(\frac{5}{16}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{16}\)
Step-by-step Solution
Detailed explanation
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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