If the probability density function of a random variable X is f(x) = (x/2) in 0 le x le 2, then P(X 1.5 | X 1) is equal to

Question

If the probability density function of a random variable X is f(x) = (x/2) in 0 le x le 2, then P(X > 1.5 | X > 1) is equal to (A) (7/16) (B) (3/4) (C) (7/12) (D) (21/64)

Tardigrade Admin · Accepted Answer

∫21.5f (x)dx, where f (x)=(x/2) ∫21.5(x/2) dx =(1/2).∫21.5xdx=(1/2)[(x2/2)]21.5 =(1/4)[4-2.25]=(1/4)×[1.75]=(175/400)=(7/16)

Q. If the probability density function of a random variable $X$ is $f (x)$ = $\frac{x}{2} in 0 \leq x \leq 2,$ then $P (X > 1.5∣ X > 1)$ is equal to

$\frac{7}{16}$

$\frac{3}{4}$

$\frac{7}{12}$

$\frac{21}{64}$

$\int_{1.5}^{2} f (x) d x,$ where $f (x) = \frac{x}{2}$
$\int_{1.5}^{2} \frac{x}{2} d x = \frac{1}{2} . \int_{1.5}^{2} x d x = \frac{1}{2} [\frac{x ^{2}}{2}]_{_{_{_{1.5}}}}^{^{^{2}}}$
$= \frac{1}{4} [4 - 2.25] = \frac{1}{4} \times [1.75] = \frac{175}{400} = \frac{7}{16}$

Q. If the probability density function of a random variable X is f(x) = 2x​in0≤x≤2, then P(X>1.5∣X>1) is equal to

167​

43​

127​

6421​