Our previous lesson introduced the concepts of random variables and defined the two categories, recall:
An easy way to think about it is this: outcomes for a DRV are countable and outcomes for a CRV generally arise from variables we measure.
Now that we know the difference between a DRV and CRV, we want to take a closer look at DRVs, and in particular, we want to look at the distribution of their probabilities.
For the probability distribution of a discrete random variable, each outcome is assigned a probability.
For example, let's say we toss a coin twice and we're interested in how many tails we see.
Firstly, we can define our DRV: Let $X$X be the number of tails in a two-coin toss.
We then know that $X$X can take on the values of $0$0, $1$1 or $2$2.
The probability distribution for $X$X will show us the probabilities for each of these outcomes. An easy way to do that for this example is to look at a tree diagram.
We'll now examine our tree diagram and tabulate the probabilities.
$x$x | $0$0 | $1$1 | $2$2 |
---|---|---|---|
$P(X=x)$P(X=x) | $\frac{1}{4}$14 | $\frac{2}{4}$24 | $\frac{1}{4}$14 |
What we've just created is the probability distribution for the random variable $X$X.
A probability distribution consists of all the outcomes together with the probability of each outcome. There are a few ways to represent a probability distribution for a DRV:
The following table gives an example of the same probability distribution given in three different representations:
Table | Function | Graph | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|
|
$P(x)=\frac{x}{10}$P(x)=x10 for $x=1,2,3,4$x=1,2,3,4. |
A probability distribution for a discrete random variable must adhere to the following conditions:
Consider the following table.
$x$x | $2$2 | $4$4 | $5$5 | $6$6 | $7$7 |
---|---|---|---|---|---|
$P$P$($($X=x$X=x$)$) | $0.1$0.1 | $0.25$0.25 | $0.3$0.3 | $0.15$0.15 | $0.2$0.2 |
Identify which conditions for a discrete probability distribution are evident in the table.
Select all that apply.
$X$X is discrete numerical
$\Sigma P\left(x\right)=1$ΣP(x)=1
$0\le P$0≤P$\left(x\right)\le1$(x)≤1
None of the conditions
Therefore, does this table represent a discrete probability distribution?
No
Yes
A special case of discrete probability distributions is the uniform distribution, where each outcome of the experiment is equally likely.
For example, when you roll a dice once, the probabilities of rolling a $1$1, $2$2, $3$3, $4$4, $5$5 or $6$6 are all equally likely, or uniform.
The graph of this probability distribution is shown below and we can observe each column is the same uniform height.
Consider the following graph.
Identify which conditions for a discrete probability distribution are evident in the graph.
Select all that apply.
$X$X is discrete numerical
$\Sigma P\left(x\right)=1$ΣP(x)=1
$0\le P$0≤P$\left(x\right)\le1$(x)≤1
None of the conditions
Therefore, does this graph represent a discrete probability distribution?
Yes
No
Consider the following function $P(X=x)=-\frac{x}{3}$P(X=x)=−x3 for $x=1,2,3$x=1,2,3.
Complete the table.
$x$x | $1$1 | $2$2 | $3$3 |
---|---|---|---|
$P(X=x)$P(X=x) | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Does this table represent a discrete probability distribution?
Yes
No
The probability function for a discrete random variable is given by:
$P$P$($($X=x$X=x$)$) | $=$= | $k\left(9-x\right)$k(9−x); $x=4,5,6,7,8$x=4,5,6,7,8 | |
$0$0, for all other values of $x$x |
Determine the exact value of $k$k.
Hence complete the table of values.
$x$x | $4$4 | $5$5 | $6$6 | $7$7 | $8$8 |
---|---|---|---|---|---|
$P$P$($($X=x$X=x$)$) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Calculate $P$P$($($X<7$X<7$)$).
Calculate $P$P$($($X\ge6$X≥6$)$).
Calculate $P$P$($($X<6$X<6$\cup$∪$X>7$X>7$)$).
Calculate $P$P$($($X\ge5$X≥5$|$|$X\le7$X≤7$)$).
The probability function for a uniform discrete random variable is given below:
$P$P$($($X=x$X=x$)$) | $=$= | $k$k; $x=1,2,3,4$x=1,2,3,4 | |
$0$0, for all other values of $x$x |
Determine the value of $k$k.
Calculate $P$P$($($X<3$X<3$)$).
Calculate $P$P$($($X\ge2$X≥2 $|$| $X<4$X<4$)$).
Determine $m$m such that $P$P$($($X\ge m$X≥m$)=0.75$)=0.75.