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A random variate $X$ takes the values $0,1,2,3$ and its mean is 1.3. If $P(X=3)=2 P(X=1)$ and $P(X=2)=0.3$, then $P(X=0)$ is equal to :
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Verified Answer
The correct answer is:
0.4
Given that,
Mean $=\Sigma X_k P(X=k)=1.3$
$X_0 P(X=0)+X_1 P(X=1)+X_2 P(X=2)+X_3 P(X=3)=1.3$
$\Rightarrow 0 \cdot P(X=0)+1 \cdot P(X=1)+2 P(X=2)$
$+3 \cdot P(X=3)=1.3$
$\Rightarrow P(X=1)+2(0.3)+3.2 P(X=1)=1.3$
$\Rightarrow 7 P(X=1)=0.7 \Rightarrow P(X=1)=0.1$
Now, $P(X=3) 2 P(X=1)=2(0.1)=0.2$
Also,
$P(X=0)+P(X=1)+P(X=2)+P(X=3)=1$
$\Rightarrow P(X=0)+0.2+0.3+0.2=1$
$\therefore P(X=0)=1-0.6=0.4$
Mean $=\Sigma X_k P(X=k)=1.3$
$X_0 P(X=0)+X_1 P(X=1)+X_2 P(X=2)+X_3 P(X=3)=1.3$
$\Rightarrow 0 \cdot P(X=0)+1 \cdot P(X=1)+2 P(X=2)$
$+3 \cdot P(X=3)=1.3$
$\Rightarrow P(X=1)+2(0.3)+3.2 P(X=1)=1.3$
$\Rightarrow 7 P(X=1)=0.7 \Rightarrow P(X=1)=0.1$
Now, $P(X=3) 2 P(X=1)=2(0.1)=0.2$
Also,
$P(X=0)+P(X=1)+P(X=2)+P(X=3)=1$
$\Rightarrow P(X=0)+0.2+0.3+0.2=1$
$\therefore P(X=0)=1-0.6=0.4$
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