Q.3.77 - A First Course in Probability

Question

Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes $1,2 or 3,$. 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 $1$.

Step-by-Step Solution

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 1: Given Information (part a)

The first trial results in outcome $1$

Step 2: 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 $1 or 2$, 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: 

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

Step 3: 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. 

Step 4: 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:

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

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

$=\frac{1}{3}$

Step 5: Final Answer (part a)

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

Step 6: Given Information (part b)

The first two trials both result in outcome $1$ 

Step 7: 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

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

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

$=\frac{1}{6}$

Step 8: Final Answer (part b)

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