(A) The sum of all probabilities in a probability distribution must be equal to $1$. $\sum P(X=x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1

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(A) The sum of all probabilities in a probability distribution must be equal to $1$. $\sum P(X=x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$ $10k^2 + 9k = 1$ $10k^2 + 9k - 1 = 0$ $(10k - 1)(k + 1) = 0$ Since $k > 0$,we have $k = \frac{1}{10}$. We need to find $P(X \geqslant 6) = P(X=6) + P(X=7)$. $P(X=6) = 2k^2 = 2(\frac{1}{10})^2 = \frac{2}{100}$. $P(X=7) = 7k^2 + k = 7(\frac{1}{10})^2 + \frac{1}{10} = \frac{7}{100} + \frac{10}{100} = \frac{17}{100}$. Therefore,$P(X \geqslant 6) = \frac{2}{100} + \frac{17}{100} = \frac{19}{100}$.

Class 12 Mathematics Probability Probability distribution

Medium

If a random variable $X$ has the following probability distribution of $X$:

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

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

Then $P(X \geqslant 6) = $

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A
$\frac{19}{100}$
B
$\frac{81}{100}$
C
$\frac{9}{100}$
D
$\frac{91}{100}$

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If $f(x) = \frac{x}{8}$ for $0 < x < 4$ and $f(x) = 0$ otherwise,is the probability density function (p.d.f.) of a continuous random variable $X$,and $F(x)$ is the cumulative distribution function (c.d.f.) associated with $f(x)$,then find $F(0.5)$.

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The following table shows the probability distribution of smart phones sold in a shop per day:
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Then $E(x) = ?$

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Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

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$A$ box contains $6$ pens,$2$ of which are defective. Two pens are taken randomly from the box. If random variable $x$ represents the number of defective pens obtained,then the standard deviation of $x$ is:

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If the function $f$ defined by $f(x) = \begin{cases} K(x-x^2) & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}$ is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is

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Suppose three coins are tossed simultaneously. If $X$ denotes the number of heads,then the probability distribution of $X$ is:

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$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

If a random variable $X$ has the following probability distribution of $X$: $X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$ Then $P(X \geqslant 6) = $

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If $f(x) = \frac{x}{8}$ for $0 < x < 4$ and $f(x) = 0$ otherwise,is the probability density function (p.d.f.) of a continuous random variable $X$,and $F(x)$ is the cumulative distribution function (c.d.f.) associated with $f(x)$,then find $F(0.5)$.

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If a random variable $X$ has the following probability distribution of $X$:

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

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

Then $P(X \geqslant 6) = $

The following table shows the probability distribution of smart phones sold in a shop per day:
Number of smart phones $(x)$ $0$ $1$ $2$ $3$ $4$ $5$
Probability $(P(x))$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$

Then $E(x) = ?$