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The 铿乺st trial results in outcome$1$

Events:

T - $3$is the last of three outcomes to appear

A - the sequence starts with$1$

B - the sequence start with $1,1$

If $3$is the last outcome to appear, the first can be either $1or2$, and those are equally likely:

$P(A鈭�T)=\frac{1}{2}$

This can be more strictly proven by the following method, here used to P(B/T)

Start with the definition:

$P(B鈭�T)=\frac{P\left(BT\right)}{P\left(T\right)}$

Now to calculate P(B T) and P(T), show them as the union of simpler events whose probability can be calculated as the sum of a geometric sequence.

P(T)

T happens if and only if one of these events occurs: the sequence starts with one or more of outcome $1$, then an outcome $2$, or the other way around, start with $2$s and then a $1$. These two are mutually exclusive thanks to the different first outcome, therefore:

$P\left(T\right)=P("1鈥�12")+P("2鈥�21")$

And each of the events on the right-hand side is again a union of mutually exclusive events, depending on how many repetitions of the first outcome appear before the other outcome occurs.

Using independence to show the probability of an intersection as a product of probabilities we obtain

$P\left(T\right)=\underset{n=1,2,鈥�}{鈭�}P("1"{)}^{n}P("2^{鈥测赌�})+\underset{n=1,2,鈥�}{鈭�}P{("2^{鈥测赌�})}^{n}P("1")$

$=2\underset{n=1,2,鈥�}{鈭�}{\left(\frac{1}{3}\right)}^n\frac{1}{3}$

The right-hand side is a geometric sequence, and it's known

$\underset{n=0,1,2鈥�}{鈭�}{q}^k=\frac{1}{1鈭�q}$

Therefore:

$P\left(T\right)=2{\left(\frac{1}{3}\right)}^2\underset{n=1,2,鈥�}{鈭�}{\left(\frac{1}{3}\right)}^{n鈭�1}$

$=2{\left(\frac{1}{3}\right)}^2\frac{1}{1鈭�\frac{1}{3}}$

$=\frac{1}{3}$

Probability given that outcome 3 is the last one to occur that the sequence starts with one outcome 1 is $P(A\backslash T)=1/2$.

The 铿乺st two trials both result in outcome$1$

P(BT)

This event can be easily shown as the union of events whose probabilities are a geometric sequence.The first two trials are $1,1,$the $K=0,1,2,...$outcomes $1$again,and then a $2$.

$={\left(\frac{1}{3}\right)}^3\underset{n=0,1,2,鈥�}{鈭�}{\left(\frac{1}{3}\right)}^n$

$={\left(\frac{1}{3}\right)}^3\underset{n=0,1,2,鈥�}{鈭�}{\left(\frac{1}{3}\right)}^n$

$=\frac{1}{3^3}鈰�\frac{1}{1鈭�\frac{1}{3}}$

$=\frac{1}{18}$

Returning to equation (1), the definition

$P(B鈭�T)=\frac{P\left(BT\right)}{P\left(T\right)}$

$=\frac{\frac{1}{18}}{\frac{1}{3}}$

$=\frac{1}{6}$

The 铿乺st two trials both result in outcome $1$is$\frac{1}{6}$