3.1: Basic Probabilities and Probability Distributions; Three Ways to Define Probabilities
Toss a thumb tack one time. Do you think the tack will land with the point up or the point down?
We cannot predict which way the tack will land before we toss it. Sometimes it will land with the point up and other times it will land with the point down. Tossing a tack is a random experiment since we cannot predict what the outcome will be. We do know that there are only two possible outcomes for each trial of the experiment: lands points up or lands point down. If we repeat the experiment of tossing the tack many times we might be able to guess how likely it is that the tack will land point up.
- A random experiment is an activity or an observation whose outcome cannot be predicted ahead of time
- A trial is one repetition of a random experiment.
- The sample space is the set of all possible outcomes for a random experiment.
- An event is a subset of the sample space.
Do you think chances of the tack landing point up and the tack landing point down are the same? This is an example where your intuition may be wrong. Having only two possible outcomes does not mean each outcome has a 50/50 chance of happening. In fact we are going to see that the probability of the tack landing point up is about 66%.
To begin to answer this question, toss a tack 10 times. For each toss record whether the tack lands point up or point down.
| Trial | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Up/Down | Up | Down | Up | Up | Down | Up | Up | Down | Up | Down |
The random experiment here is tossing the tack one time. The possible outcomes for the experiment are that the tack lands point up or the tack lands point down so the sample space is S = {point up, point down}. We are interested in the event E that the tack lands point up, E = {point up}.
Based on our data we would say that the tack landed point up six out of 10 times. The fraction, , is called the relative frequency. Since we would guess that probability of the tack landing point up is about 60%.
Let’s repeat the experiment by tossing the tack ten more times.
| Trial | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Up/Down | Up | Up | Up | Down | Down | Up | Up | Down | Up | Up |
This time the tack landed point up seven out of 10 times or 70% of the time. If we tossed the tack another 10 times we might get a different result again. The probability of the tack landing point up refers to what happens when we toss the tack many, many times. Let’s toss the tack 150 times and count the number of times it lands point up. Along the way we will look at the proportion of landing point up.
| Trials | Number Up | Total Number of Up | Total Number of Trials | Proportion |
|---|---|---|---|---|
| 0-10 | 6 | 6 | 10 | 6/10=0.60 |
| 11-20 | 7 | 13 | 20 | 13/20=0.65 |
| 21-30 | 8 | 21 | 30 | 21/30=0.70 |
| 31-40 | 6 | 27 | 40 | 27/40=0.68 |
| 41-50 | 8 | 35 | 50 | 35/50=0.70 |
| 51-60 | 8 | 43 | 60 | 43/60=0.72 |
| 61-70 | 4 | 47 | 70 | 47/70=0.67 |
| 71-80 | 7 | 54 | 80 | 54/80=0.68 |
| 81-90 | 5 | 59 | 90 | 59/90=0.66 |
| 91-100 | 6 | 65 | 100 | 65/100=0.65 |
| 101-110 | 10 | 75 | 110 | 75/110=0.68 |
| 111-120 | 5 | 80 | 120 | 80/120=0.67 |
| 121-130 | 8 | 88 | 130 | 88/130=0.68 |
| 131-140 | 4 | 92 | 140 | 92/140=0.66 |
| 141-150 | 7 | 99 | 150 | 99/150=0.66 |
When we have a small number of trials the proportion varies quite a bit. As we start to have more trials the proportion still varies but not by as much. It appears that the proportion is around 0.66 or 66%. We would have to do about 100,000 trials to get a better approximation of the actual probability of the tack landing point up.
The tack landed point up 99 out of 150 trials. The probability P of event E is written as:
\[P(E)=\frac{\# \text { of trials with point up }}{\text { total number of trials }}=\frac{99}{150} \approx 0.66 \nonumber \]
We would say the probability that the tack lands point up is about 66%.
Equally Likely Outcomes
In some experiments all the outcomes have the same chance of happening. If we roll a fair die the chances are the same for rolling a two or rolling a five. If we draw a single card from a well shuffled deck of cards, each card has the same chance of being selected. We call outcomes like these equally likely. Drawing names from a hat or drawing straws are other examples of equally likely outcomes. The tack tossing example did not have equally likely outcomes since the probability of the tack landing point up is different than the probability of the tack landing point down.
An experiment has equally likely outcomes if every outcome has the same probability of occurring.
For equally likely outcomes, the probability of outcome A , P(A) , is:
\[P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }}. \nonumber \]
Round Off Rule : Give probabilities as a fraction or as a decimal number rounded to three decimal places.
Draw a single card from a well shuffled deck of 52 cards. Each card has the same chance of being drawn so we have equally likely outcomes. Find the following probabilities:
- P (card is red)
P (card is red) = \(\dfrac{ \text {number of red cards}}{\text { total number of cards }} = \dfrac{26}{52} = \dfrac{1}{2}\)
The probability that the card is red is \(\dfrac{1}{2}\).
- P (card is a heart)
P (card is a heart) = \(\dfrac{ \text {number of hearts}}{\text { total number of cards }} = \dfrac{13}{52} = \dfrac{1}{4}\)
The probability that the card is a heart is \(\dfrac{1}{4}\).
- P (card is a red 5)
P (card is a red 5) = \(\dfrac{ \text {number of red fives}}{\text { total number of cards }} = \dfrac{2}{52} = \dfrac{1}{26}\)
The probability that the card is a red five is \(\dfrac{1}{26}\).
Roll a fair die one time. The sample space is S = {1, 2, 3, 4, 5, 6}. Find the following probabilities.
- P (roll a four)
P (roll a four) = \(\dfrac{ \text {number of ways to roll a four}}{\text { total number of ways to roll a die }} = \dfrac{1}{6}\)
The probability of rolling a four is \(\dfrac{1}{6}\).
- P (roll an odd number)
The event roll an odd number is E = {1, 3, 5}.
P (roll an odd number) = \(\dfrac{ \text {number of ways to roll an odd number}}{\text { total number of ways to roll a die }} = \dfrac{3}{6} = \dfrac{1}{2}\)
The probability of rolling an odd number is \(\dfrac{1}{2}\).
- P (roll a number less than five)
The event roll a number less than five is F = {1, 2, 3, 4}.
P (roll a number less than five) = \(\dfrac{ \text {number of ways to roll number less than five}}{\text { total number of ways to roll a die }} = \dfrac{4}{6} = \dfrac{2}{3}\)
The probability of rolling a number less than five is \(\dfrac{2}{3}\).
A small bookcase contains five math, three English and seven science books. A book is chosen at random. What is the probability that a math book is chosen?
Solution
Since the book is chosen at random each book has the same chance of being chosen and we have equally likely events.
\[P(\text { math book })=\frac{\text { number of ways to choose a math book }}{\text { total number of books }}=\frac{5}{15}=\frac{1}{3} \nonumber \]
The probability a math book was chosen is \(\dfrac{1}{3}\).
Three Ways of Finding Probabilities
There are three ways to find probabilities. In the tack tossing example we calculated the probability of the tack landing point up by doing an experiment and recording the outcomes. This was an example of an empirical probability. The probability of getting a red jack in a card game or rolling a five with a fair die can be calculated from mathematical formulas. These are examples of theoretical probabilities. The third type of probability is a subjective probability. Saying that there is an 80% chance that you will go to the beach this weekend is a subjective probability. It is based on experience or guessing.
- A theoretical probability is based on a mathematical model where all outcomes are equally likely to occur.
- An empirical probability is based on an experiment or observation and is the relative frequency of the event occurring.
- A subjective probability is an estimate (a guess) based on experience or intuition.
Complements
If there is a 75% chance of rain today, what are the chances it will not rain? We know that there are only two possibilities. It will either rain or it will not rain. Because the sum of the probabilities for all the outcomes in the sample space must be 100% or 1.00, we know that
P (will rain) + P (will not rain) = 100%.
Rearranging this we see that
P (will not rain) = 100% - P (will rain) = 100% - 75% = 25%.
The events E = {will rain} and F = {will not rain} are called complements.
The complement of event E , denoted by \(\overline{E}\), is the set of outcomes in the sample space that are not in the event E . The probability of \(\overline{E}\) is given by \(P(\overline{E}) = 1- P(E)\).
Draw a single card from a well shuffled deck of 52 cards.
- Look at the suit of the card. Here the sample space S = {spades, clubs, hearts, diamonds}. If event E = {spades} the complement \(\overline{E}\) = {clubs, hearts, diamonds}.
- Look at the value of the cards. Here the sample space is S = {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K}. If the event E = {the number is less than 7} = {A, 2, 3, 4, 5, 6} the complement \(\overline{E}\) = {7, 8, 9, 10, J, Q, K}.
A train arrives on time at a particular station 85% of the time. Does this mean that the train is late 15% of the time? The answer is no. The complement of E = {on time} is not \(\overline{E}\) = {late}. There is a third possibility. The train could be early. The sample space is S = {on time, early, late} so the complement of E = {on time} is \(\overline{E}\) = {early or late}. Based on the given information we cannot find P (late) but we can find P (early or late) = 15%.
Impossible Events and Certain Events
Recall that \(P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }}\). What does it mean if we say the probability of the event is zero? \(P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }} = 0\). The only way for a fraction to equal zero is when the numerator is zero. This means there is no way for event A to occur. A probability of zero means that the event is impossible.
What does it mean if we say the probability of an event is one? \(P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }} = 1\). The only way for a fraction to equal one is if the numerator and denominator are the same. The number of ways for A to occur is the same as the number of outcomes. There are no outcomes where A does not occur. A probability of 1 means that the event always happens.
- \(P(A) = 0\) means that A is impossible
- \(P(A) = 1\) means that A is certain
Probability Distributions
A probability distribution (probability space) is a sample space paired with the probabilities for each outcome in the sample space. If we toss a fair coin and see which side lands up, there are two outcomes, heads and tails. Since the coin is fair these are equally likely outcomes and have the same probabilities. The probability distribution would be P (heads) = 1/2 and P (tails) = 1/2. This is often written in table form:
| Outcome | Heads | Tails |
|---|---|---|
| Probability | 1/2 | 1/2 |
| A probability distribution for an experiment is a list of all the possible outcomes and their corresponding probabilities. |
Example \(\PageIndex{6}\): Probabilities for the Sum of Two Fair Dice
In probability problems when we roll two dice, it is helpful to think of the dice as being different colors. Let’s assume that one die is red and the other die is green. We consider getting a three on the red die and a five on the green die different than getting a five on the red die and a three on the green die. In other words, when we list the outcomes the order matters. The possible outcomes of rolling two dice and looking at the sum are given in Table \(\PageIndex{8}\).
| 1+1 = 2 | 1+2 = 3 | 1+3 = 4 | 1+4 = 5 | 1+5 = 6 | 1+6 = 7 |
| 2+1 = 3 | 2+2 = 4 | 2+3 = 5 | 2+4 = 6 | 2+5 = 7 | 2+6 = 8 |
| 3+1 = 4 | 3+2 = 5 | 3+3 = 6 | 3+4 = 7 | 3+5 = 8 | 3+6 = 9 |
| 4+1 = 5 | 4+2 = 6 | 4+3 = 7 | 4+4 = 8 | 4+5 = 9 | 4+6 = 10 |
| 5+1 = 6 | 5+2 = 7 | 5+3 = 8 | 5+4 = 9 | 5+5 = 10 | 5+6 = 11 |
| 6+1 = 7 | 6+2 = 8 | 6+3 = 9 | 6+4 = 10 | 6+5 = 11 | 6+6 = 12 |
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Probability | |||||||||||
|
Reduced Probability |
Are the following valid probability distributions?
| Outcome | A | B | C | D | E |
|---|---|---|---|---|---|
| Probability |
This is a valid probability distribution. All the probabilities are between zero and one inclusive and the sum of the probabilities is 1.00.
| Outcome | A | B | C | D | E | F |
|---|---|---|---|---|---|---|
| Probability | 0.45 | 0.80 | -0.20 | -0.35 | 0.10 | 0.20 |
This is not a valid probability distribution. The sum of the probabilities is 1.00, but some of the probabilities are not between zero and one, inclusive.
| Outcome | A | B | C | D |
|---|---|---|---|---|
| Probability | 0.30 | 0.20 | 0.40 | 0.25 |
This is not a valid probability distribution. All of the probabilities are between zero and one, inclusive, but the sum of the probabilities is 1.15 not 1.00.
Odds
Probabilities are always numbers between zero and one. Many people are not comfortable working with such small values. Another way of describing the likelihood of an event happening is to use the ratio of how often it happens to how often it does not happen. The ratio is called the odds of the event happening. There are two types of odds, odds for and odds against. Casinos, race tracks and other types of gambling usually state the odds against an event happening.
If the probability of an event E is P(E) , then the odds for event E , O(E) , are given by:
\(O(E) = \dfrac{P(E)}{P(\overline{E})}\) OR \(O(E) = \dfrac{\text{number of ways for E to occur}}{\text{number of ways for E to not occur}}\)
Also, the odds against event E , are given by:
\(O(\overline{E}) = \dfrac{P(\overline{E})}{P(E)}\) OR \(O(E) = \dfrac{\text{number of ways for E to not occur}}{\text{number of ways for E to occur}}\)
A single card is drawn from a well shuffled deck of 52 cards. Find the odds that the card is a red eight.
There are two red eights in the deck.
\(P(\text{red eight}) = \frac{2}{52} = \frac{1}{26}\).
\(P(\text{not a red eight}) = \frac{50}{52} = \frac{25}{26}\).
\(O(\text{red eight}) = \dfrac{P(\text{red eight})}{P(\text{not a red eight})} = \dfrac{\frac{1}{26}}{\frac{25}{26}} = \frac{1}{\cancel{26}} \cdot \frac{\cancel{26}}{25}= \frac{1}{25}\)
The odds of drawing a red eight are 1 to 25. This can also be written as 1:25.
Note: Do not write odds as a decimal or a percent.
Many roulette wheels have slots numbered 0, 00, and 1 through 36. The slots numbered 0 and 00 are green. The even numbered slots are red and the odd numbered slots are black. The game is played by spinning the wheel one direction and rolling a marble around the outer edge the other direction. Players bet on which slot the marble will fall into. What are the odds the marble will land in a red slot?
Solution
There are 38 slots in all. The slots 2, 4, 6, …, 36 are red so there are 18 red slots. The other 20 slots are not red.
\(P(\text{red}) = \frac{18}{38} = \frac{9}{19}\).
\(P(\text{not red}) = 1 - \frac{9}{19} = \frac{19}{19} - \frac{9}{19} = \frac{10}{19}\).
\(O(\text{red}) = \dfrac{P(\text{red})}{P(\text{not red})} = \dfrac{\frac{9}{19}}{\frac{10}{19}} = \frac{9}{\cancel{19}} \cdot \frac{\cancel{19}}{10}= \frac{9}{10}\)
The odds of the marble landing in a red slot are 9 to 10. This can also be written as 9:10.
Two fair dice are tossed and the sum is recorded. Find the odds against rolling a sum of nine.
Solution
The event E , roll a sum of nine is: E = {(3, 6), (4, 5), (5, 4), (6, 3)}
There are 36 ways to roll two dice and four ways to roll a sum of nine. That means there are 32 ways to roll a sum that is not nine.
\(P(\text{sum is nine}) = \frac{4}{36} = \frac{1}{9}\).
\(P(\text{sum is not nine}) = \frac{32}{36} = \frac{8}{9}\).
\(O(\text{against sum is nine}) = \dfrac{P(\text{sum is not nine})}{P(\text{sum is nine})} = \dfrac{\frac{8}{9}}{\frac{1}{9}} = \frac{8}{\cancel{9}} \cdot \frac{\cancel{9}}{1}= \frac{8}{1}\)
The odds against rolling a sum of nine are 8 to 1 or 8:1.
We can also find the probability of an event happening based on the odds for the event. Saying that the odds of an event are 3 to 5 means that the event happens three times for every five times it does not happen. If we add up the possibilities of both we get a sum of eight. So the event happens about three out of every eight times. We would say the probability is 3/8.
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If the odds favoring event E are a to b , then: \(P(E) = \dfrac{a}{a+b}\) and \(P(\overline{E}) = \dfrac{b}{a+b}\). |
A local little league baseball team is going to a tournament. The odds of the team winning the tournament are 3 to 7. Find the probability of the team winning the tournament.
Solution
\(P(\text{winning}) = \dfrac{3}{3+7} = \dfrac{3}{10} = 0.3\)