# [SOLVED]-aux08.probability distribution find a and b

#### karush

##### Well-known member
A discrete random variable $X$ has a probability distribution as shown in the table below.
1103
(a) Find the value of $a+b$
if the sum of $E[X] = 1$ then $1-0.1-0.3 =0.6=a+b$
(b) Given $E[X]=1.5$ find the value of a and b
why would this be $1.5$

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#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
(b) Given $$\displaystyle E[X]=1.5$$, find the value of $$\displaystyle a$$ and $$\displaystyle b$$

why would this be $$\displaystyle 1.5$$
I am not sure I understand the question. The fact that $E[X]=1.5$ is given. You don't question why $P(X=0)=0.1$, do you?

Just write what $E[X]$ is by definition for this particular $X$, and you'll get a second linear equation in $a$ and $b$ in addition to $a+b=0.6$.

#### karush

##### Well-known member
I am not sure I understand the question. The fact that $E[X]=1.5$ is given. You don't question why $P(X=0)=0.1$, do you?

Just write what $E[X]$ is by definition for this particular $X$, and you'll get a second linear equation in $a$ and $b$ in addition to $a+b=0.6$.
well that gives $$\displaystyle a+b =1.1$$ but still how do we get values for $$\displaystyle a$$ and $$\displaystyle b$$

#### MarkFL

Staff member
...
if the sum of $$\displaystyle E[X] =$$1 then $$\displaystyle 1-0.1-0.3 =0.6=a+b$$

(b) Given $$\displaystyle E[X]=1.5$$, find the value of $$\displaystyle a$$ and $$\displaystyle b$$

why would this be $$\displaystyle 1.5$$

$$\displaystyle P(0)+P(1)+P(2)+P(3)=1$$

and this gives, as you correctly found, $$\displaystyle a+b=0.6$$

Now, you also know:

$$\displaystyle 0.1\cdot0+a\cdot1+0.3\cdot2+b\cdot3=E[X]=1.5$$

or

$$\displaystyle a+3b=0.9$$

So, you have the linear system:

$$\displaystyle a+b=0.6$$

$$\displaystyle a+3b=0.9$$

I would suggest beginning by subtracting the first equation from the second to eliminate $a$...

#### karush

##### Well-known member

$$\displaystyle P(0)+P(1)+P(2)+P(3)=1$$

and this gives, as you correctly found, $$\displaystyle a+b=0.6$$

Now, you also know:

$$\displaystyle 0.1\cdot0+a\cdot1+0.3\cdot2+b\cdot3=E[X]=1.5$$

or

$$\displaystyle a+3b=0.9$$

So, you have the linear system:

$$\displaystyle a+b=0.6$$

$$\displaystyle a+3b=0.9$$

I would suggest beginning by subtracting the first equation from the second to eliminate $a$...
ok got.. $$\displaystyle a=0.45$$ and $$\displaystyle b=0.15$$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
the original table is lost
I was wondering if is possible to reconstruct it from the comments
From post #4 it seems that the distribution is the following.
$$\displaystyle \begin{array}{c|c|c|c|c} x & 0 & 1 & 2 & 3\\ \hline P(X=x) & 0.1 & a & 0.3 & b \end{array}$$

#### karush

##### Well-known member
From post #4 it seems that the distribution is the following.
$$\displaystyle \begin{array}{c|c|c|c|c} x & 0 & 1 & 2 & 3\\ \hline P(X=x) & 0.1 & a & 0.3 & b \end{array}$$
much mahalo