Q80 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
A random variable $X$ has the following distribution
\begin{array}{lllllllll}
\hline \begin{array}{l}
Values of \\
X(x)
\end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hlineP(X=x) \quad 0 & k & 2 k & 2 k & 3 k & k^2 & 2 k^2 & 7 k^2+k \\
\hline
\end{array}
- A$\frac{9}{10}$
- Correct answerB$\left(\frac{9}{10}\right)^2$
- C$\left(\frac{3}{10}\right)$
- D$\frac{1}{10}$
Solution
$$
\begin{aligned}
& \text { Since, } \Sigma P\left(X=x_i\right)=1 \\
& \Rightarrow 0+K+2 K+2 K+3 K+K^2+2 K^2 \\
& +7 K^2+K=1 \\
& \Rightarrow \quad 9 K+10 K^2=1 \\
& \Rightarrow \quad K=\frac{1}{10}, \quad[\because K>0] \\
&
\end{aligned}
$$
So,
$$
\begin{aligned}
& P(0 < K < 6)=K+2 K+2 K+3 K+K^2 \\
& =8 K+K^2=\frac{8}{10}+\frac{1}{100}=\frac{81}{100}=\left(\frac{9}{10}\right)^2
\end{aligned}
$$
Hence, the correct answer is option: 'B'.