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A string consisting of $A$s, $B$s and $C$s is chosen uniformly at random from the set $\{BBBBB, ABBBC, AACCC, ABBCC, BBBBC \}$.

Let $X$ be the number of $A$s in the string.

Give the probability distribution of $X$.

Do i assume $X$ as the number of $A$'s in the set which will be $X= \{0,1,2\}$ . If so how do i continue as the question is quite new to me .

Zubzub
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Dave Kumar
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    Your assuming is a good start. For $p(X)$ there is 2 strings out of 5 where $X=0$, 2 out of 5 where $X=1$ and one case with $X=2$. What are the probabilities those cases to occur? – M.P May 09 '17 at 09:15

2 Answers2

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So we have that the probability that $X = 0$ occurs if $BBBBB$ or $BBBBC$ is chosen. Hence $X=0\ w.p.\ 2/5$.

Then $X=1$ occurs if $ABBBC$ or $ABBCC$ is chosen. Hence $X=1\ w.p.\ 2/5$.

I let you finish !

Zubzub
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Here $\Omega:=\{BBBBB, ABBBC, AACCC, ABBCC, BBBBC \}$.

It is equipped with $\sigma$-algebra $\wp(\Omega)$ and with a probability measure $P$ defined by $P(E)=\frac15|E|$.

This together makes $\langle\Omega,\wp(\Omega),P\rangle$ a probability space.

Then $X:\Omega\to\mathbb R$ is prescribed by: $$\omega\mapsto\text{number of }A\text{s in string }\omega$$

$X$ will take values in $\{0,1,2\}$.

Here e.g. $\{X=1\}=\{\omega\in\Omega\mid X(\omega)=1\}=\{ABBBC,ABBCC\}$ so that $$P(\{X=1\})=P(\{ABBBC,ABBCC\})=\frac25$$

It is common to abbreviate $P(\{X=1\})$ by $P(X=1)$.

Another way to write the result is: $P_X(1)=\frac25$.

drhab
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