alleryFor p.d.f. $f(x) = x^2/3$ for $-1 < x < 2$, the probability $P[0 < X \leq 1]$ equals:
A$8/9$
B$1/9$
C$1/3$
D$2/9$
Answer & Solution
Correct answer: B. $1/9$
$P[0 < X \leq 1] = \int_0^1 x^2/3\,dx = [x^3/9]_0^1 = 1/9$.
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