Probability

Section 4.B Probability

Determine sample and event spaces.
Calculate probabilities using event and sample spaces.
Calculate probabilities from one- and two-way tables.
Calculate probabilities using Venn diagrams.

Subsection 4.B.1 Introduction

One story about how probability theory was developed is that a gambler wanted to know when to bet more and when to bet less. He talked to a couple of friends of his who happened to be mathematicians. Their names were Pierre de Fermat and Blaise Pascal. Since then many other mathematicians have worked to develop probability theory.

Understanding probabilities are important in life. Examples of mundane questions that probability can answer for you are if you need to carry an umbrella or wear a heavy coat on a given day. More important questions that probability can help with are your chances that the car you are buying will need more maintenance, your chances of winning the lottery or your chances of being in a car accident. The chance of you winning the lottery is very small, yet many people will spend the money on lottery tickets. Yet, if instead they saved the money that they spend on the lottery, they would have more money. In general, events that have a low probability (under \(5\%\)) are unlikely to occur. Whereas, if an event has a high probability of happening (over \(80\%\)), then there is a good chance that the event will happen. This section will present some of the theory that you need to help make a determination of whether an event is likely to happen or not.

Subsection 4.B.2 Experiments and Sample Spaces

First you need some definitions.

Experiment: An activity that has specific results that can occur, but it is unknown which results will occur.
Outcomes: The results of an experiment.
Event: A set \(A\) of certain outcomes of an experiment that you want to have happen.
Sample Space: collection of all possible outcomes of the experiment. Usually denoted as \(SS\text{.}\)
Event space: The set of outcomes that make up an event. The symbol is usually a capital letter.

If all of the outcomes in a sample space are equally likely to happen, then the probability of an event \(A\) is calculated as follows.

The probability of event \(A\) happening is

\begin{gather} P(A) = \dfrac{\text{# of outcomes in event space } A}{\text{# of outcomes in sample space }SS }=\dfrac{n(A)}{n(SS)}\tag{4.B.1} \end{gather}

Example 4.B.1. One Die.

Consider the experiment of rolling an ordinary 6-sided die (and observing the number of dots showing up on the roll).

The sample space has \(6\) outcomes, one for each side of the die:

\begin{gather*} SS=\{1,2,3,4,5,6 \} \end{gather*}

(a) Determine the probability of rolling a \(6\text{.}\) First write the event space, \(A\text{:}\)

\begin{align*} A=\{6\}\amp\amp\amp\text{The event \(A\) contains \(1\) outcome.} \end{align*}

Since \(P(A)=\dfrac{\text{# of outcomes in event space \(A\)}}{\text{# of outcomes in sample space \(SS\)}}\text{,}\)

\begin{align*} P(A)=\dfrac{n(A)}{n(SS)}=\dfrac{1}{6}\approx 0.1667\amp\amp\amp\text{Our Solution (a):} \end{align*}

The probability of rolling a \(6\) is \(\dfrac{1}{6}\approx 0.1667\) or about \(16.67\%\checkmark\text{.}\)

(b) Determine the probability of rolling an even number. First write the event space, \(B\text{:}\)

\begin{align*} B=\{2,4,6 \}\amp\amp\amp\text{The event \(B\) contains \(3\) outcomes, the even numbers on the die.}\\ P(B) =\dfrac{n(B)}{n(SS)}=\dfrac{3}{6}=\dfrac{1}{2}=0.5\amp\amp\amp\text{Our Solution (b):} \end{align*}

The probability of rolling an even number is \(\dfrac{1}{2}=0.5\) or \(50\%\checkmark\text{.}\)

Example 4.B.2. Three Children--Gender sample space.

A family has three children. Write a sample space (for the possible genders of children in the family).

Let \(B\) represent a boy. Let \(G\) represent a girl. First determine the number of outcomes in the sample space.

There are two choices for each child's gender. We create a table as in Set Section 4.A.

Child 1	Child 2	Child 3
\(2\)	\(2\)	\(2\)

\begin{align*} 2\cdot 2 \cdot 2\amp\amp\amp \text{Multiply the possibilities using the \textbf{Multiplication Axiom}}\\ 8 \amp\amp\amp \text{There are \(8\) elements in the sample space.} \end{align*}

Illustrate the possibilities with a tree diagram. The tree has \(8\) endpoints as expected, one for each element in the sample space.

The possibility \(BGB\text{,}\) for example, indicates that the first born is a boy, the second born a girl, and the third a boy.

\begin{align*} SS=\{BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG \} \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

Example 4.B.3. Three Children--Probability of at least 2 girls.

A family has three children. Determine the probability of the family having at least two girls.

Recall from Example Example 4.B.2 the sample space \(SS\) has \(8\) outcomes:

\begin{gather*} SS=\{BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG\} \end{gather*}

To determine the probability of the family having at least two girls, first write the event space, \(A\) at least two girls:

\begin{align*} A=\{BGG, GBG, GGB, GGG \} \amp\amp\amp \text{The event \(A\) contains \(4\) outcomes}\\ P(A) = \dfrac{n(A)}{n(SS)}=\dfrac{4}{8}=\dfrac{1}{2}=0.5\amp\amp\amp \text{Our Solution:} \end{align*}

The probability of the family having at least two girls is \(\dfrac{1}{2}\) or \(50\%.\checkmark\)

Example 4.B.4. Two Dice Sample Space.

Two fair dice are rolled. Write the sample space.

We assume one of the dice is red, and the other green. We create a table as before:

	\(\color{green}{\text{Green}}\)
\(\color{red}{\text{Red}}\)	\(\color{green}{1}\)	\(\color{green}{2}\)	\(\color{green}{3}\)	\(\color{green}{4}\)	\(\color{green}{5}\)	\(\color{green}{6}\)
\(\color{red}{1}\)	\((\color{red}{1},\color{green}{1})\)	\((\color{red}{1},\color{green}{2})\)	\((\color{red}{1},\color{green}{3})\)	\((\color{red}{1},\color{green}{4})\)	\((\color{red}{1},\color{green}{5})\)	\((\color{red}{1},\color{green}{6})\)
\(\color{red}{2}\)	\((\color{red}{2},\color{green}{1})\)	\((\color{red}{2},\color{green}{2})\)	\((\color{red}{2},\color{green}{3})\)	\((\color{red}{2},\color{green}{4})\)	\((\color{red}{2},\color{green}{5})\)	\((\color{red}{2},\color{green}{6})\)
\(\color{red}{3}\)	\((\color{red}{3},\color{green}{1})\)	\((\color{red}{3},\color{green}{2})\)	\((\color{red}{3},\color{green}{3})\)	\((\color{red}{3},\color{green}{4})\)	\((\color{red}{3},\color{green}{5})\)	\((\color{red}{3},\color{green}{6})\)
\(\color{red}{4}\)	\((\color{red}{4},\color{green}{1})\)	\((\color{red}{4},\color{green}{2})\)	\((\color{red}{4},\color{green}{3})\)	\((\color{red}{4},\color{green}{4})\)	\((\color{red}{4},\color{green}{5})\)	\((\color{red}{4},\color{green}{6})\)
\(\color{red}{5}\)	\((\color{red}{5},\color{green}{1})\)	\((\color{red}{5},\color{green}{2})\)	\((\color{red}{5},\color{green}{3})\)	\((\color{red}{5},\color{green}{4})\)	\((\color{red}{5},\color{green}{5})\)	\((\color{red}{5},\color{green}{6})\)
\(\color{red}{6}\)	\((\color{red}{6},\color{green}{1})\)	\((\color{red}{6},\color{green}{2})\)	\((\color{red}{6},\color{green}{3})\)	\((\color{red}{6},\color{green}{4})\)	\((\color{red}{6},\color{green}{5})\)	\((\color{red}{6},\color{green}{6})\)

\begin{align*} SS=\{(1,1), (1,2),...,(6,6) \} \amp\amp\amp \text{Our Solution:} \end{align*}

The sample space is the \(36=6\times 6\) outcomes listed in the table above. \(\checkmark\)

Example 4.B.5. Two Dice-Probabilities of certain sums.

Two fair dice are rolled. Determine the following probabilities.

In Example Example 4.B.4 we, we determined the sample space \(SS\) has the \(36\) outcomes listed.

(a) Determine the probability of rolling a sum of \(6\text{.}\) First write \(D\) for the sum of the dots showing. Then write the event space \((D=6)\) a sum of \(6\text{:}\)

\begin{align*} (D=6) = \{(\color{red}{1},\color{green}{5}), (\color{red}{2},\color{green}{4}), (\color{red}{3},\color{green}{3}), (\color{red}{4},\color{green}{2}), (\color{red}{5},\color{green}{1}) \amp\amp\amp \text{The event \((D=6)\) contains \(5\) outcomes}\\ \text{\(P(D=6) =\dfrac{n(D)}{n(SS)}=\dfrac{5}{36}\approx 0.1389\)} \amp\amp\amp \text{Our Solution (a):} \end{align*}

The probability of a sum of \(6\) on the two dice is \(\dfrac{5}{36}\) or about \(13.89\%.\checkmark\)

(b) Determine the probability of rolling a sum less than or equal to \(4\text{.}\) First write the event space, \((D \leq 4)\text{:}\)

\begin{align*} (D \leq 4)= \{(\color{red}{1},\color{green}{1}), (\color{red}{1},\color{green}{2}), (\color{red}{1},\color{green}{3}), (\color{red}{2},\color{green}{1}), (\color{red}{2},\color{green}{2}), (\color{red}{3},\color{green}{1}) \} \amp\amp\amp \text{The event \((D \leq 4)\) contains \(6\) outcomes}\\ P(D \leq 4) =\dfrac{n(D \leq 4)}{n(SS)}=\dfrac{6}{36}=\dfrac{1}{6}\approx 0.1667 \amp\amp\amp \text{Our Solution (b):} \end{align*}

The probability of a sum less than or equal to \(4\) on the two dice is \(\dfrac{1}{6}\) or about \(16.67\%.\checkmark\)

Example 4.B.6. Cards--Sample space and probabilities.

Suppose you conduct an experiment where you pull a card from a standard deck of cards. Recall this deck has \(4\) suits (black spades \(S\text{,}\) black clubs \(C\text{,}\) red diamonds \(D\text{,}\) and red hearts \(H\)) each with \(13\) cards (numbers \(2\) through \(10\text{,}\) jack \(J\text{,}\) queen \(Q\text{,}\) king \(K\text{,}\) and ace \(A\)). We will generally consider the face cards to be the jacks, hearts, queens, kings, and aces.

\begin{align*} SS=\amp\\ \amp\{2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AS,\\ \amp2C, 3C, 4C, 5C, 6C, 7C, 8C, 9C, 10C, JC, QC, KC, AC,\\ \amp\color{red}{2D}, \color{red}{3D}, \color{red}{4D}, \color{red}{5D}, \color{red}{6D}, \color{red}{7D}, \color{red}{8D}, \color{red}{9D}, \color{red}{10D}, \color{red}{JD}, \color{red}{QD}, \color{red}{KD}, \color{red}{AD},\\ \amp\color{red}{2H}, \color{red}{3H}, \color{red}{4H}, \color{red}{5H}, \color{red}{6H}, \color{red}{7H}, \color{red}{8H}, \color{red}{9H}, \color{red}{10H}, \color{red}{JH}, \color{red}{QH}, \color{red}{KH}, \color{red}{AH}\} \end{align*}

The sample space has \(52\) outcomes, one for each card in the deck. Determine the following probabilities.

(a) Determine the probability of selecting a spade. First write the event space, \(Spade\text{.}\)

\begin{align*} \text{\(Spade=\) \(\{2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AS \}\)} \amp\amp\amp \text{The event \(Spade\) contains \(13\) outcomes.}\\ \text{\(P(Spade) = \dfrac{n(Spade)}{n(SS)}=\dfrac{13}{52}=\dfrac{1}{4}=0.25\)} \amp\amp\amp \text{Our Solution (a):} \end{align*}

The probability of selecting a spade is \(\dfrac{13}{52}\) or \(25\%.\checkmark\)

(b) Determine the probability of selecting a jack. First write the event space, \(Jack\text{.}\)

\begin{align*} \text{\(Jack=\{JS, JC, \color{red}{JH}, \color{red}{JD} \}\)} \amp\amp\amp \text{The event \(Jack\) contains \(4\) outcomes.}\\ \text{\(P(Jack) =\dfrac{n(Jack)}{n(SS)}=\dfrac{4}{52}=\dfrac{1}{13}\approx 0.07692\)} \amp\amp\amp \text{Our Solution (b):} \end{align*}

The probability of selecting a jack is \(\dfrac{4}{52}\) or about \(7.69\%.\checkmark\)

(c) Determine the probability of selecting a club or a face card. First write the event space, \(Club \cup Face\text{.}\)

\begin{align*} \amp Club \cup Face=\amp\amp\\ \amp\{2C, 3C, 4C, 5C, 6C, 7C, 8C, 9C, 10C, JC, QC, KC, AC,\amp\amp\\ \amp JS, QS, KS, AS, \color{red}{JD}, \color{red}{QD}, \color{red}{KD}, \color{red}{AD}, \color{red}{JH}, \color{red}{QH}, \color{red}{KH}, \color{red}{AH}\} \amp\amp \text{The event \(Club \cup Face\) contains \(25\) outcomes.}\\ \amp P (Club \cup Face)= \dfrac{n(Club \cup Face)}{n(SS)}=\dfrac{25}{52} \approx 0.480769 \amp\amp \text{Our Solution (c):} \end{align*}

The probability of selecting a club or a face card is \(\dfrac{25}{52}\) or about \(48.08\%.\checkmark\)

It is also possible to compute probabilities based on the number of times an event \(A\) occurred.

The probability of event \(A\) is

\begin{gather} P(A) = \dfrac{n(A)}{\text{Total # of outcomes}}\tag{4.B.2} \end{gather}

Example 4.B.7. One-Way Table Probabilities.

A bag of chocolate candies contained the following distribution of colors.

Blue \((B)\)	Purple \((P)\)	Green \((G)\)	Orange \((O)\)	Red \((R)\)	Yellow \((Y)\)	Total
\(481\)	\(371\)	\(483\)	\(544\)	\(372\)	\(369\)	\(2620\)

If all of the candies are placed in a bowl and one is selected at random, determine the probability of selecting the following candies.

(a) Determine the probability of selecting a blue candy. Label the event space \(B\) for blue.

\begin{align*} P(B) = \dfrac{n(B)}{\text{Total #}}=\dfrac{481}{2620}\approx 0.1835 \amp\amp\amp \text{Our Solution (a):} \end{align*}

The probability of selecting a blue candy is \(\dfrac{481}{2620}\) or about \(18.35\%.\checkmark\)

(b) Determine the probability of selecting a purple or yellow candy. Label the event spaces: \(P\) for purple and \(Y\) for yellow.

\begin{align*} \text{Then the event purple OR yellow is denoted \(P\cup Y\).} \amp\amp\amp n(P\cup Y)=371+369=740\\ P(P\cup Y) = \dfrac{n(P\cup Y)}{\text{Total #}}=\dfrac{740}{2620}\approx 0.2824 \amp\amp\amp \text{Our Solution (b):} \end{align*}

The probability of selecting a purple or yellow candy is \(\dfrac{740}{2620}\) or about \(28.24\%.\checkmark\)

Example 4.B.8. Two-Way Table Probabilities.

The number of people who survived the Titanic based on class and gender is in the below table ("Encyclopedia Titanica," 2013). Suppose a person is picked at random from the survivors.

	Gender
Class	Female (\(F\))	Male (\(M)\)	Total
\(1st\)	\(134\)	\(59\)	\(193\)
\(2nd\)	\(94\)	\(25\)	\(119\)
\(3rd\)	\(80\)	\(58\)	\(138\)
Total	\(308\)	\(142\)	\(450\)

Determine the following probabilities.

(a) Determine the probability that a survivor was female. Label the event space \(F\) for female.

\begin{align*} P(F) = \dfrac{n(F)}{\text{Total #}}=\dfrac{308}{450}\approx 0.6844\amp\amp\amp \text{Our Solution (a):} \end{align*}

The probability that a survivor was female is \(\dfrac{308}{450}\) or about \(68.44\%.\checkmark\)

(b) Determine the probability that a survivor was female and in the first class. Label the event space \(F\cap 1st\) for female AND in the first class.

\begin{align*} n(F\cap 1st)=134 \amp\amp\amp \text{The number of survivors in both the}\\ \amp\amp\amp\text{\(F\) column and \(1st\) row.}\\ P(F\cap 1st) = \dfrac{n(F\cap 1st)}{\text{Total #}}=\dfrac{134}{450}\approx 0.2978 \amp\amp\amp \text{Our Solution (b):} \end{align*}

The probability that a survivor was a first-class female is \(\dfrac{134}{450}\) or about \(29.78\%.\checkmark\)

(c) Determine the probability that a survivor was female or in the 1st class. Label the event space \(F\cup 1st\) for female OR in the first class.

\begin{align*} n(F\cup 1st)=134+94+80+59=367 \amp\amp\amp \text{The number of survivors in either the}\\ \amp\amp\amp\text{ \(F\) column or 1st row.}\\ P(F\cup 1st) = \dfrac{n(F\cup 1st)}{\text{Total #}}=\dfrac{367}{450}\approx 0.8156\amp\amp\amp \text{Our Solution (c):} \end{align*}

The probability that a survivor was female OR in first class is \(\dfrac{367}{450}\) or about \(81.56\%.\checkmark\)

It may be useful to calculate probabilities using a Venn diagram.

Example 4.B.9. Venn Diagram Probabilities.

Recall in Example Example 4.A.16 we considered a survey of \(100\) people in California indicating that \(60\) people had visited Disneyland, \(15\) had visited Knott's Berry Farm, and \(6\) had visited both. Determine the following probabilities using the Venn diagram.

Let set \(D\) represent those surveyed who have visited Disneyland.

Let set \(K\) represent those surveyed who have visited Knott's Berry Farm.

We completed the Venn diagram at the right.

(a) Determine the probability that a person surveyed went to Disneyland. Label the event space \(D\) for Disneyland.

\begin{align*} n(D)=54+6=60 \amp\amp\amp \text{The number of people surveyed who went to Disneyland.}\\ P(D) = \dfrac{n(D)}{\text{Total #}}=\dfrac{60}{100}= 0.6 \amp\amp\amp \text{Our Solution (a):} \end{align*}

The probability that a person surveyed went to Disneyland is \(\dfrac{60}{100}\) or \(60\%.\checkmark\)

(b) Determine the probability that a person surveyed went to both parks. Label the event space \(D\cap K\) for Disneyland AND Knott's.

\begin{align*} n(D\cap K)=6 \amp\amp\amp \text{The number of people surveyed who went to both parks.}\\ P(D\cap K) = \dfrac{n(D\cap K)}{\text{Total #}}=\dfrac{6}{100}= 0.06 \amp\amp\amp \text{Our Solution (b):} \end{align*}

The probability that a person surveyed went to both parks is \(\dfrac{6}{100}\) or \(6\%.\checkmark\)

(c) Determine the probability that a person surveyed went to either Knott's or Disneyland (or both). Label the event space \(D\cup K\) for Disneyland OR Knott's.

\begin{align*} n(D\cup K)=54+6+9=69 \amp\amp\amp \text{The number of people surveyed who went to either park (or both).}\\ P(D\cup K) =\dfrac{69}{100}= 0.69 \amp\amp\amp \text{Our Solution (c):} \end{align*}

The probability that a person surveyed went to either park is \(\dfrac{69}{100}\) or \(69\%.\checkmark\text{.}\)