Discrete random variables

Worked example 1 Define the variable before its distribution

Two fair coins are tossed independently. Let \(X\) be the number of heads in these two tosses. Find the probability distribution of \(X\).

Solution

Write H for heads and T for tails. Each fair coin gives either result with probability \(\frac12\). Independence means one coin’s result does not affect the other, so we multiply probabilities: each ordered outcome below has probability \(\frac12\times\frac12=\frac14\):

Group outcomes that give the same value of \(X\):

Two outcomes give one head, so \(\operatorname{P}(X=1)=\frac12\). The three values of \(X\) are not equally likely.

Worked example 2 Find a missing constant in a table

Let \(X\) be the number of messages received in a particular ten-minute interval. Its model is

Find \(k\) and verify that the completed table is a valid probability distribution.

Solution

The probabilities must add to 1: \[\begin{aligned} k+2k+3k+4k&=1,\\ 10k&=1,\\ k&=\boxed{\frac1{10}}. \end{aligned}\] The completed probabilities are \(0.1,0.2,0.3,0.4\). Each lies between 0 and 1, and their sum is 1. Both conditions hold.

Worked example 3 Find a constant and construct the distribution

A spinner has three unequal sectors labelled 1, 2 and 3. Let \(X\) be the number obtained on one spin. Its probability distribution is \[\operatorname{P}(X=x)=k(x+1),\qquad x\in\{1,2,3\},\] with probability zero for all other values. Find \(k\) and represent the distribution as a table and a probability bar chart.

Solution

First substitute the allowed values: \[\begin{aligned} \operatorname{P}(X=1)&=k(1+1)=2k,\\ \operatorname{P}(X=2)&=k(2+1)=3k,\\ \operatorname{P}(X=3)&=k(3+1)=4k. \end{aligned}\] Now make the total probability equal to 1: \[2k+3k+4k=1\quad\Longrightarrow\quad 9k=1\quad\Longrightarrow\quad k=\boxed{\frac19}.\]

All entries lie between 0 and 1, and \(\frac29+\frac39+\frac49=1\).

Worked example 4 Inequalities and impossible values

Let \(X\) be the change in a player’s score, in points, after one round:

Find:

1. \(\operatorname{P}(X>1)\);

2. \(\operatorname{P}(X\geq1)\);

3. \(\operatorname{P}(X\leq0)\) using the complement;

4. \(\operatorname{P}(X=2)\).

Solution

1. Only 3 is greater than 1: \(\operatorname{P}(X>1)=\operatorname{P}(X=3)=\boxed{\frac3{10}}\).

2. Both 1 and 3 are at least 1: \[\operatorname{P}(X\geq1)=\frac4{10}+\frac3{10}=\boxed{\frac7{10}}.\]

3. The positive values are 1 and 3, so \[\operatorname{P}(X\leq0)=1-\operatorname{P}(X>0)=1-\left(\frac4{10}+\frac3{10}\right)=\boxed{\frac3{10}}.\]

4. The value 2 is not in the distribution: \(\boxed{\operatorname{P}(X=2)=0}\).

Worked example 5 Calculate and interpret a mean

Let \(X\) be the number of tokens in one randomly selected envelope, with

Find \(\operatorname{E}(X)\) and explain what it means.

Solution

Keep each value matched with its probability:

Add the products in the bottom row: \[\operatorname{E}(X)=0\left(\frac15\right)+2\left(\frac12\right)+5\left(\frac3{10}\right)=\boxed{\frac52\text{ tokens}}.\] Over many independent selections with this same distribution, the average number of tokens per envelope will tend towards 2.5. For example, counts near 200, 500 and 300 in 1000 selections would give an average near \(2.5\); actual counts fluctuate.

Worked example 6 From a daily mean to an expected total

Let \(X\) be the number of express deliveries made by a small shop in one day:

1. Find and interpret \(\operatorname{E}(X)\).

2. Find the expected number of express deliveries over 30 days, assuming this model applies each day.

Solution

1. Multiply the values by their probabilities and add: \[\begin{aligned} \operatorname{E}(X)&=0(0.10)+1(0.25)+2(0.40)+3(0.25)\\ &=\boxed{1.8\text{ deliveries}}. \end{aligned}\] Over many days, the shop averages about 1.8 express deliveries per day. A single day’s count is still a whole number.

2. The expected total is \(30(1.8)=\boxed{54\text{ deliveries}}\). The actual total could be different.

Worked example 7 Use a given mean to find probabilities

Let \(X\) be the number of points earned in one round of a game:

Given that \(\operatorname{E}(X)=\frac52\), find \(p\) and \(q\).

Solution

There are two unknown probabilities, so use two facts: \[\begin{aligned} p+\frac12+q&=1 &&\text{(total probability)},\\ 0p+2\left(\frac12\right)+6q&=\frac52 &&\text{(expected value)}. \end{aligned}\] From the second equation, \[1+6q=\frac52\quad\Longrightarrow\quad6q=\frac32\quad\Longrightarrow\quad q=\frac14.\] Then \(p=1-\frac12-\frac14=\frac14\). The probabilities \(\frac14,\frac12,\frac14\) lie between 0 and 1 and sum to 1. Check the mean: \[0\left(\frac14\right)+2\left(\frac12\right)+6\left(\frac14\right)=\frac52,\] which agrees with the given value.

Worked example 8 A prize does not tell the whole story

A player pays €2 to select one prize envelope. Let \(R\) be the prize money received, in euros, on this play:

1. Find the expected prize money, \(\operatorname{E}(R)\).

2. Define the player’s net-gain distribution and find its mean.

3. Find the probability that the player makes a profit. Is the game fair?

Solution

1. \(\operatorname{E}(R)=0(\frac15)+2(\frac12)+5(\frac3{10})=\boxed{\text{€}2.50}\).

2. Let \(G\) be the player’s net gain, in euros, on this play. Since \(G=R-2\), subtract 2 from each prize; the corresponding probabilities stay the same:

   \(\text{Prize }r\)	\(0\)	\(2\)	\(5\)
   \(\text{Net gain }g\)	\(-2\)	\(0\)	\(3\)
   \(\operatorname{P}(G=g)\)	\(\frac15\)	\(\frac12\)	\(\frac3{10}\)

   \[\operatorname{E}(G)=(-2)\left(\frac15\right)+0\left(\frac12\right)+3\left(\frac3{10}\right)=\boxed{\text{€}0.50}.\]

3. A profit requires \(G>0\), which occurs only when the prize is €5: \[\operatorname{P}(G>0)=\boxed{\frac3{10}}.\] The game is not fair: \(\operatorname{E}(G)>0\). The player gains an average of €0.50 per play in the long run, even though a profit occurs on only 30% of plays.

Worked example 9 Find the fair entry fee

In the envelope game, the prize \(R\) is €0, €2 or €5, with probabilities \(\frac15,\frac12,\frac3{10}\). Find the entry fee \(c\) euros that makes the game fair.

Solution

Let \(G\) be the player’s net gain, in euros, on one play. Since \(G=R-c\), its values are \(-c\), \(2-c\) and \(5-c\), with the same probabilities as the prizes. \[\begin{aligned} \operatorname{E}(G)&=(-c)\left(\frac15\right)+(2-c)\left(\frac12\right)+(5-c)\left(\frac3{10}\right)\\ &=\frac52-c\left(\frac15+\frac12+\frac3{10}\right)\\ &=\frac52-c. \end{aligned}\] For fairness, \(\operatorname{E}(G)=0\), so \(\frac52-c=0\) and \(\boxed{c=2.50}\). The fair entry fee is €2.50.

Worked example 10 Find the prize that makes a game fair

A spinner has eight equal sectors. One sector wins a prize of €\(w\); the other seven pay nothing. It costs €3 to play. Find \(w\) for a fair game.

Solution

Let \(G\) be the player’s net gain, in euros, on one spin. The possible net gains are

Set the expected net gain equal to zero: \[\begin{aligned} (w-3)\left(\frac18\right)+(-3)\left(\frac78\right)&=0,\\ w-3-21&=0,\\ w&=\boxed{24}. \end{aligned}\] The prize must be €24. Check: the net gains are €21 and \(-\text{€}3\), and \(21(\frac18)-3(\frac78)=0\).