Question

Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes $1,2or3,$. Given that outcome $3$is the last of the three outcomes to occur, find the conditional probability that

(a) the first trial results in outcome $1$;

(b) the first two trials both result in outcome $1$.

Short answer

Probability given that outcome 3 is the last one to occur that the sequence starts with one outcome 1 is $P(A\backslash T)=1/2$, with two outcomes $1-P(B\backslash T)=1/6$

Step-by-step solution

Given Information (part a)

The first trial results in outcome$1$

Explanation (part a)

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:

$P(A\mid T)=\frac{1}{2}$

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

Start with the definition:

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

Explanation (part a)

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:

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

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.

Explanation (part a)

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

$P\left(T\right)=\sum _{n=1,2,\dots }P("1"{)}^{n}P("2^{\prime \prime })+\sum _{n=1,2,\dots }P{("2^{\prime \prime })}^{n}P("1")$

$=2\sum _{n=1,2,\dots }{\left(\frac{1}{3}\right)}^n\frac{1}{3}$

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

$\sum _{n=0,1,2\dots }{q}^k=\frac{1}{1-q}$

Therefore:

$P\left(T\right)=2{\left(\frac{1}{3}\right)}^2\sum _{n=1,2,\dots }{\left(\frac{1}{3}\right)}^{n-1}$

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

$=\frac{1}{3}$

Final Answer (part a)

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

Given Information (part b)

The first two trials both result in outcome$1$

Explanation (part b)

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 $2$.

$={\left(\frac{1}{3}\right)}^3\sum _{n=0,1,2,\dots }{\left(\frac{1}{3}\right)}^n$

$={\left(\frac{1}{3}\right)}^3\sum _{n=0,1,2,\dots }{\left(\frac{1}{3}\right)}^n$

$=\frac{1}{3^3}\cdot \frac{1}{1-\frac{1}{3}}$

$=\frac{1}{18}$

Returning to equation (1), the definition

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

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

$=\frac{1}{6}$

Final Answer (part b)

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