-aux08.probability distribution find a and b

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Discussion Overview

The discussion revolves around finding the values of variables \(a\) and \(b\) in a probability distribution of a discrete random variable \(X\). Participants explore the implications of the expected value \(E[X]\) being set to both 1 and 1.5, and how these conditions relate to the probabilities assigned to different outcomes of \(X\).

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose that if \(E[X] = 1\), then \(a + b = 0.6\) based on the probabilities summing to 1.
  • Others argue that with \(E[X] = 1.5\), a second equation can be derived, leading to a system of equations: \(a + b = 0.6\) and \(a + 3b = 0.9\).
  • A later reply questions how to derive specific values for \(a\) and \(b\) from the equations, leading to a proposed solution of \(a = 0.45\) and \(b = 0.15\).
  • Some participants express uncertainty about the original probability distribution table, which is referenced but not provided, leading to discussions about reconstructing it from the comments.

Areas of Agreement / Disagreement

Participants generally agree on the equations derived from the expected value conditions, but there is no consensus on the specific values of \(a\) and \(b\) until later posts. The original probability distribution table remains unclear, leading to further speculation.

Contextual Notes

The discussion lacks the original table of probabilities, which is essential for fully understanding the context of the problem. Additionally, there are unresolved assumptions regarding the values of \(a\) and \(b\) and how they relate to the probabilities assigned to \(X\).

karush
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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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karush said:
(b) Given $$E[X]=1.5$$, find the value of $$a$$ and $$b$$

why would this be $$1.5$$
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 $a+b=0.6$.
 
Evgeny.Makarov said:
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 $$a+b =1.1$$ but still how do we get values for $$a$$ and $$b$$
 
karush said:
...
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 $$1.5$$

It is instead:

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

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

Now, you also know:

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

or

$$a+3b=0.9$$

So, you have the linear system:

$$a+b=0.6$$

$$a+3b=0.9$$

I would suggest beginning by subtracting the first equation from the second to eliminate $a$...
 
MarkFL said:
It is instead:

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

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

Now, you also know:

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

or

$$a+3b=0.9$$

So, you have the linear system:

$$a+b=0.6$$

$$a+3b=0.9$$

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

ok got.. $$a=0.45$$ and $$b=0.15$$
 
the original table is lost
I was wondering if is possible to reconstruct it from the comments
mahalo
SSCtw.png
 
karush said:
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.
[math]
\begin{array}{c|c|c|c|c}
x & 0 & 1 & 2 & 3\\
\hline
P(X=x) & 0.1 & a & 0.3 & b
\end{array}
[/math]
 
Evgeny.Makarov said:
From post #4 it seems that the distribution is the following.
[math]
\begin{array}{c|c|c|c|c}
x & 0 & 1 & 2 & 3\\
\hline
P(X=x) & 0.1 & a & 0.3 & b
\end{array}
[/math]
much mahalo
 

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