The range of a random variable $X$ is $\{0,1,2\}$. If $P(X=0)$ $=3 \mathrm{C}^3, \mathrm{P}(\mathrm{X}=1)=4…
The range of a random variable $X$ is $\{0,1,2\}$. If $P(X=0)$ $=3 \mathrm{C}^3, \mathrm{P}(\mathrm{X}=1)=4 \mathrm{C}-10 \mathrm{C}^2$ and $\mathrm{P}(\mathrm{X}=2)=5 \mathrm{C}-1$, then the value of $\mathrm{C}$ is
$\frac{2}{3}$
$\frac{1}{3}$
. $\frac{5}{3}$
$\frac{4}{3}$
Solution
$
\begin{aligned}
& \text { } \because P(X=0)+P(X=1)+P(X=2)=1 \\
& \Rightarrow 3 c^3+4 c-10 c^2+5 c-1=1 \\
& \Rightarrow 3 c^3+4 c-10 c^2+5 c-2=0 \\
& \Rightarrow(c-1)(3 c-1)(c-2)=0 \\
& \Rightarrow c=\frac{1}{3}, 1,2 .
\end{aligned}
$
When $c=2$, Then $P(X=2)=10-1=9>1$
$
\therefore \quad c \neq 2
$
When $c=1$, Then $P(X=2)=5-1=4>1$
$
\therefore \quad c \neq 1
$
So, $c=\frac{1}{3}$