If the random variable X has the following probability distribution:
|
x |
-1 |
0 |
1 |
2 |
|
P(X = x) |
k |
2k |
3k |
$\frac{k}{2}$ |
then P(X ≤ 0) is equal to:
Answer & explanation
Correct answer: option 1
Given
$X : -1,\,0,\,1,\,2$
$P(X=x) : k,\,2k,\,3k,\,\frac{k}{2}$
Total probability $=1$
$k+2k+3k+\frac{k}{2}=1$
$6k+\frac{k}{2}=1$
$\frac{13k}{2}=1$
$k=\frac{2}{13}$
Now
$P(X\le0)=P(X=-1)+P(X=0)$
$=k+2k=3k$
$=3\left(\frac{2}{13}\right)=\frac{6}{13}$
$P(X\le0)=\frac{6}{13}$.