The probability distribution of a random vari | vedclass

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(C) For a probability distribution, the sum of all probabilities must be $1$:$\sum P(X=x) = a + a + a + b + b + 0.3 = 1$$3a + 2b + 0.3 = 1$$3a + 2b =

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$0.1, 0.2$

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(C) For a probability distribution, the sum of all probabilities must be $1$:$\sum P(X=x) = a + a + a + b + b + 0.3 = 1$$3a + 2b + 0.3 = 1$$3a + 2b = 0.7$  --- $(i)$The mean of a random variable $X$ is given by $E(X) = \sum x_i P(x_i) = 4.2$:$1(a) + 2(a) + 3(a) + 4(b) + 5(b) + 6(0.3) = 4.2$$a + 2a + 3a + 4b + 5b + 1.8 = 4.2$$6a + 9b = 4.2 - 1.8$$6a + 9b = 2.4$Dividing by $3$, we get:$2a + 3b = 0.8$  --- $(ii)$Now, solve the system of linear equations $(i)$ and $(ii)$:Multiply $(i)$ by $3$ and $(ii)$ by $2$:$9a + 6b = 2.1$  --- $(iii)$$4a + 6b = 1.6$  --- $(iv)$Subtract $(iv)$ from $(iii)$:$5a = 0.5 \Rightarrow a = 0.1$Substitute $a = 0.1$ into $(i)$:$3(0.1) + 2b = 0.7$$0.3 + 2b = 0.7$$2b = 0.4 \Rightarrow b = 0.2$Thus, $a = 0.1$ and $b = 0.2$.

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	$a$	$a$	$a$	$b$	$b$	$0.3$

$X = x$	$-1$	$0$	$1$	$2$
$F(X = x)$	$0.3$	$0.7$	$0.8$	$1$

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

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

If the mean of $X$ is $4.2$, then $a$ and $b$ are respectively equal to:

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$A$ random variable $X$ has the following probability distribution:
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The probability distribution of a random variable $X$ is given below:$x$$1$$2$$3$$4$$5$$6$$P(X=x)$$a$$a$$a$$b$$b$$0.3$If the mean of $X$ is $4.2$, then $a$ and $b$ are respectively equal to: