Using Sample spaces to determine Probabilities

We can determine probabilities if we can completely specify an outcome space. In doing this, it is important to distinguish between the physical outcomes of an experiment on the one hand and the feature of interest on the other.

Consider an experiment where a coin is flipped twice in succession. The feature of interest may be the number of heads that occur. What is the probability that exactly one head is flipped?

It would be a mistake to think that since either $0$0, $1$1, or $2$2 heads may be flipped that the probability of one head occurring is $\frac{1}{3}$13​. Instead we should list all the possible outcomes of the trial:

$\left\{HH,HT,TH,TT\right\}${HH,HT,TH,TT}.

The number of occurrences of the feature of interest (exactly one head) is $2$2 out of the $4$4 possible outcomes, for an overall probability of $\frac{2}{4}=\frac{1}{2}$24​=12​.

We can now generalise using a random variable. If $X$X is the number of heads in a single trial, we can write 

$P\left(X=1\right)=\frac{1}{2}$P(X=1)=12​,

which we read as "the probability that there is exactly $1$1 head is $\frac{1}{2}$12​".

Similarly,

$P\left(X=0\right)=\frac{1}{4}$P(X=0)=14​ and $P\left(X=2\right)=\frac{1}{4}$P(X=2)=14​,

indicating that the probability that there are $0$0 heads is $\frac{1}{4}$14​, and the same for $2$2 heads.

In general, the probability of the occurrence of each value depends on the number of times the defining feature of the random variable occurs in the sample space.

Imagine a machine randomly choosing pairs $\left(x,y\right)$(x,y)of numbers from the set $\left\{1,2,3,4\right\}${1,2,3,4}. The outcome space is this set:

$$$\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(2,1\right),\left(2,3\right),\left(2,4\right),$(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),

$\left(3,1\right),\left(3,2\right),\left(3,4\right),\left(4,1\right),\left(4,2\right),\left(4,3\right)$(3,1),(3,2),(3,4),(4,1),(4,2),(4,3).$$

Suppose the feature of interest is the sum of the numbers in each pair. That is, we define a random variable $A=x+y$A=x+y. We might ask: What is the probability that the sum $A=6$A=6? This result occurs twice out of twelve outcomes in the space, and so we conclude that $P(A=6)=\frac{1}{6}$P(A=6)=16​.

A useful device for ensuring that all the outcomes of an outcome space have been listed is the tree diagram. The following diagram shows how the experiment described above could be represented.

Example

With the outcome space of the experiment described above, we define a random variable $B$B that has the value $0$0 if the sum of the two numbers is even and the value $1$1 if the sum is odd. What is the probability that $B=1$B=1, that is, that the sum will be odd?

The outcomes are: 

$\left\{12,13,14,21,23,24,31,32,34,41,42,43\right\}${12,13,14,21,23,24,31,32,34,41,42,43}

We perform the further operation of summing each pair of numbers and assigning the values 0 and 1 according to whether the sums are even or odd. This gives the following corresponding set of values of the random variable $B$B:

$\left\{1,0,1,1,1,0,0,1,1,1,0,1\right\}${1,0,1,1,1,0,0,1,1,1,0,1}

Of the 12 outcomes, 8 result in an odd sum. So, $P(B=1)=\frac{2}{3}$P(B=1)=23​.

Worked Examples

Question 1

 

Question 2

QUEstion 3