A random variable X has the following probabi | vedclass

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(A) Step $1$: Find the value of $k$ using the property $\int_{0}^{1} f(x) dx = 1$.$\int_{0}^{1} k(x - x^2) dx = k [\frac{x^2}{2} - \frac{x^3}{3}]_{0}^

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$\frac{1}{3}$

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(A) Step $1$: Find the value of $k$ using the property $\int_{0}^{1} f(x) dx = 1$.$\int_{0}^{1} k(x - x^2) dx = k [\frac{x^2}{2} - \frac{x^3}{3}]_{0}^{1} = k(\frac{1}{2} - \frac{1}{3}) = k(\frac{1}{6}) = 1 \implies k = 6$.Step $2$: Use the condition $P(X > a) = \frac{20}{27}$.$P(X > a) = \int_{a}^{1} 6(x - x^2) dx = \frac{20}{27}$.$6 [\frac{x^2}{2} - \frac{x^3}{3}]_{a}^{1} = \frac{20}{27}$.$6 [(\frac{1}{2} - \frac{1}{3}) - (\frac{a^2}{2} - \frac{a^3}{3})] = \frac{20}{27}$.$6 [\frac{1}{6} - \frac{a^2}{2} + \frac{a^3}{3}] = \frac{20}{27}$.$1 - 3a^2 + 2a^3 = \frac{20}{27}$.$2a^3 - 3a^2 + 1 - \frac{20}{27} = 0 \implies 2a^3 - 3a^2 + \frac{7}{27} = 0$.$54a^3 - 81a^2 + 7 = 0$.Testing $a = \frac{1}{3}$: $54(\frac{1}{27}) - 81(\frac{1}{9}) + 7 = 2 - 9 + 7 = 0$.Thus,$a = \frac{1}{3}$ is the correct solution.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$k$	$3k$	$3k$	$k$

$x$	$-2$	$-1$	$3$	$4$	$6$
$P(X=x)$	$\frac{1}{5}$	$a$	$\frac{1}{3}$	$\frac{1}{5}$	$b$

$A$ random variable $X$ has the following probability density function (p.d.f.):
$f(x) = kx(1-x), 0 \leqslant x \leqslant 1$
If $P(X > a) = \frac{20}{27}$,then find the value of $a$.

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$A$ random variable $X$ has the following probability density function (p.d.f.):$f(x) = kx(1-x), 0 \leqslant x \leqslant 1$If $P(X > a) = \frac{20}{27}$,then find the value of $a$.