example

If we roll a standard 6-sided die, describe the sample space and some simple events.

Solution: 

The sample space is the set of all possible simple events: {1,2,3,4,5,6}

Some examples of simple events:

• We roll a 1
• We roll a 5

Some compound events:

• We roll a number bigger than 4
• We roll an even number

examples

If we roll a 6-sided die, calculate

1. P(rolling a 1)
2. P(rolling a number bigger than 4)

Solutions: 

Recall that the sample space is {1,2,3,4,5,6}

1. There is one outcome corresponding to “rolling a 1,” so the probability is [latex]\frac{1}{6}[/latex]
2. There are two outcomes bigger than a 4, so the probability is [latex]\frac{2}{6}=\frac{1}{3}[/latex]

Probabilities are essentially fractions, and can be reduced to lower terms like fractions.

This video describes this example and the previous one in detail.

Let’s say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet?

Solution: 

There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is [latex]\frac{14}{20}=\frac{7}{10}[/latex].
There is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn’t be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.

example

Compute the probability of randomly drawing one card from a deck and getting an Ace.

Solution: 

There are 52 cards in the deck and 4 Aces so [latex]P(Ace)=\frac{4}{52}=\frac{1}{13}\approx 0.0769[/latex]

We can also think of probabilities as percents: There is a 7.69% chance that a randomly selected card will be an Ace.

Notice that the smallest possible probability is 0 – if there are no outcomes that correspond with the event. The largest possible probability is 1 – if all possible outcomes correspond with the event.

This video demonstrates both this example and the previous cherry example on the page.

example

If you pull a random card from a deck of playing cards, what is the probability it is not a heart?

Solution: 

There are 13 hearts in the deck, so [latex]P(\text{heart})=\frac{13}{52}=\frac{1}{4}[/latex].

The probability of not drawing a heart is the complement: [latex]P(\text{not heart})=1-P(\text{heart})=1-\frac{1}{4}=\frac{3}{4}[/latex]

This situation is explained in the following video.

https://youtu.be/RnljiW6ZM6A

example

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.

Solution: 

We could list all possible outcomes:  {H1,H2,h2,H4,H5,H6,T1,T2,T3,T4,T5,T6}.

Notice there are [latex]2\cdot6=12[/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\frac{1}{12}[/latex].

example

Are these events independent?

1. A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.
2. The two events (1) “It will rain tomorrow in Houston” and (2) “It will rain tomorrow in Galveston” (a city near Houston).
3. You draw a card from a deck, then draw a second card without replacing the first.

Solutions: 

1. The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss, so these events are independent.
2. These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.
3. The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.

example

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the probability both are white?

Solution: 

The probability of choosing a white pair of socks is [latex]\frac{6}{10}[/latex].

The probability of choosing a white tee shirt is [latex]\frac{3}{7}[/latex].

The probability of both being white is [latex]\frac{6}{10}\cdot\frac{3}{7}=\frac{18}{70}=\frac{9}{35}[/latex]

Examples of joint probabilities are discussed in this video.

https://youtu.be/6F17WLp-EL8

example

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin or a 6 on the die.

Solution: 

Here, there are still 12 possible outcomes: {H1,H2,h2,H4,H5,H6,T1,T2,T3,T4,T5,T6}

By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on the die or both – we use or inclusively here (these 7 outcomes are H1, H2, h2, H4, H5, H6, T6), so the probability is [latex]\frac{7}{12}[/latex]. How could we have found this from the individual probabilities?

As we would expect, [latex]\frac{1}{2}[/latex] of these outcomes have a head, and [latex]\frac{1}{6}[/latex] of these outcomes have a 6 on the die. If we add these, [latex]\frac{1}{2}+\frac{1}{6}=\frac{6}{12}+\frac{2}{12}=\frac{8}{12}[/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is [latex]\frac{1}{12}[/latex].

If we subtract out this double count, we have the correct probability: [latex]\frac{8}{12}-\frac{1}{12}=\frac{7}{12}[/latex].

example

Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?

Solution: 

There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:

[latex]P(\text{King or Queen})=\frac{8}{52}[/latex]

Note that in this case, there are no cards that are both a Queen and a King, so [latex]P(\text{King and Queen})=0[/latex]. Using our probability rule, we could have said:

[latex]P(\text{King or Queen})=P(\text{King})+P(\text{Queen})-P(\text{King and Queen})=\frac{4}{52}+\frac{4}{52}-0=\frac{8}{52}[/latex]

See more about this example and the previous one in the following video.

https://youtu.be/klbPZeH1np4

example

Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?

Solution: 

Half the cards are red, so [latex]P(\text{red})=\frac{26}{52}[/latex]

There are four kings, so [latex]P(\text{King})=\frac{4}{52}[/latex]

There are two red kings, so [latex]P(\text{Red and King})=\frac{2}{52}[/latex]

We can then calculate

[latex]P(\text{Red or King})=P(\text{Red})+P(\text{King})-P(\text{Red and King})=\frac{26}{52}+\frac{4}{52}-\frac{2}{52}=\frac{28}{52}[/latex]

Example

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:

1. Has a red car and got a speeding ticket
2. Has a red car or got a speeding ticket.

Solution: 

We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is [latex]\frac{15}{665}\approx0.0226[/latex].

Notice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring.

We could answer this question by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 with red cars but no ticket + 45 with a ticket but no red car = 195 people. So the probability is [latex]\frac{195}{665}\approx0.2932[/latex].

We also could have found this probability by:

P(had a red car) + P(got a speeding ticket) – P(had a red car and got a speeding ticket)

= [latex]\frac{150}{665}+\frac{60}{665}-\frac{15}{665}=\frac{195}{665}[/latex].

This table example is detailed in the following explanatory video.

https://youtu.be/HWrGoM1yRaU

example

What is the probability that two cards drawn at random from a deck of playing cards will both be aces?

Solution: 

It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\frac{4}{52}\cdot\frac{4}{52}=\frac{1}{169}[/latex]. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.

Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case the “condition” is that the first card is an ace. Symbolically, we write this as:

P(ace on second draw | an ace on the first draw).

The vertical bar “|” is read as “given,” so the above expression is short for “The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw.” What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]\frac{3}{51}=\frac{1}{17}[/latex].

Thus, the probability of both cards being aces is [latex]\frac{4}{52}\cdot\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}[/latex].

example

If you pull 2 cards out of a deck, what is the probability that both are spades?

Solution: 

The probability that the first card is a spade is [latex]\frac{13}{52}[/latex].

The probability that the second card is a spade, given the first was a spade, is [latex]\frac{12}{51}[/latex], since there is one less spade in the deck, and one less total cards.

The probability that both cards are spades is [latex]\frac{13}{52}\cdot\frac{12}{51}=\frac{156}{2652}\approx0.0588[/latex]

Example

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:

1. has a speeding ticket given they have a red car
2. has a red car given they have a speeding ticket

Solutions: 

1. Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so P(ticket | red car) = [latex]\frac{15}{150}=\frac{1}{10}=0.1[/latex]
2. Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so P(red car | ticket) = [latex]\frac{15}{60}=\frac{1}{4}=0.25[/latex].

Notice from the last example that P(B | A) is not equal to P(A | B).

These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.

View more about conditional probability in the following video.

https://youtu.be/b6tstekMlb8

Example

If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?

Solution: 

You can satisfy this condition by having Case A or Case B, as follows:

Case A) you can get the Ace of Diamonds first and then a black card or

Case B) you can get a black card first and then the Ace of Diamonds.

Let’s calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is [latex]\frac{1}{52}[/latex]. The probability that the second card is black given that the first card is the Ace of Diamonds is [latex]\frac{26}{51}[/latex] because 26 of the remaining 51 cards are black. The probability is therefore [latex]\frac{1}{52}\cdot\frac{26}{51}=\frac{1}{102}[/latex].

Now for Case B: the probability that the first card is black is [latex]\frac{26}{52}=\frac{1}{2}[/latex]. The probability that the second card is the Ace of Diamonds given that the first card is black is [latex]\frac{1}{51}[/latex]. The probability of Case B is therefore [latex]\frac{1}{2}\cdot\frac{1}{51}=\frac{1}{102}[/latex], the same as the probability of Case 1.

Recall that the probability of A or B is P(A) + P(B) – P(A and B). In this problem, P(A and B) = 0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is [latex]\frac{1}{101}+\frac{1}{101}=\frac{2}{101}[/latex]. The probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck is [latex]\frac{2}{101}[/latex].

These two playing card scenarios are discussed further in the following video.

Click here to view this video.

Example

A home pregnancy test was given to women, then pregnancy was verified through blood tests.  The following table shows the home pregnancy test results.

Find

1. P (not pregnant | positive test result)
2. P (positive test result | not pregnant)

Solutions: 

1. Since we know the test result was positive, we’re limited to the 75 women in the first column, of which 5 were not pregnant. P(not pregnant | positive test result) = [latex]\frac{5}{75}\approx0.067[/latex].
2. Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test. P(positive test result | not pregnant) = [latex]\frac{5}{19}\approx0.263[/latex]

The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.

See more about this example here.

Click here to view this video.