(B) For a probability distribution,the sum of all probabilities must be $1$: $\sum P(X=x_i) = 1$ $0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$

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(B) For a probability distribution,the sum of all probabilities must be $1$: $\sum P(X=x_i) = 1$ $0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$ $10k^2 + 9k - 1 = 0$ $(10k - 1)(k + 1) = 0$ Since $P(X=x) \ge 0$,$k$ must be positive,so $k = \frac{1}{10}$. Now,we need to find $P(0 $P(0 Substituting $k = \frac{1}{10}$: $P(0 < X < 6) = 8\left(\frac{1}{10}\right) + \left(\frac{1}{10}\right)^2 = \frac{8}{10} + \frac{1}{100} = \frac{80+1}{100} = \frac{81}{100} = \left(\frac{9}{10}\right)^2$.

Class 12 Mathematics Probability Probability distribution

Medium

$A$ random variable $X$ has the following probability distribution:

$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Find the value of $P(0 < X < 6)$.

AP EAMCET 2018 All AP EAMCET

A
$\frac{9}{10}$
B
$\left(\frac{9}{10}\right)^2$
C
$\frac{3}{10}$
D
$\frac{1}{10}$

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The probability distribution of a random variable $X$ is given below:
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$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x_i)$	$\alpha$	$\alpha$	$\alpha$	$\beta$	$\beta$	$0.3$

$A$ random variable $X$ has the following probability distribution: $x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$ Find the value of $P(0 < X < 6)$.

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$A$ random variable $X$ has the following probability distribution:

$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Find the value of $P(0 < X < 6)$.

If a random variable $X$ has the following probability distribution,then the mean of $X$ is
$X = x_i$ $1$ $2$ $3$ $5$
$P(X = x_i)$ $2k^2$ $k$ $k$ $k^2$

The probability distribution of a random variable $X$ is given below:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X=x_i)$ $\alpha$ $\alpha$ $\alpha$ $\beta$ $\beta$ $0.3$

If $\mu$ and $\sigma^2$ represent the mean and variance of $X$ and $\mu=4.2$,then $\sigma^2+\mu^2=$