2.2 Notes/Examples

2.2 Summary Notes

From section 2.1, you found that it is quite straightforward to calculate probabilities for simple situations. What happens when we calculate probabilities from multiple events? For example, suppose you roll a single die and then flip a coin. What are the chances that the die comes up with a 5 and the coin gives you a heads? A situation that asks you to calculate probabilities for a situation that involves two or more events or steps is called a compound event. We will try to find out how to handle these types of situations by examining several situations and then making a conclusion.

Example 1

Suppose a single die is rolled and a coin is flipped. What is the probability that the die comes up with a 5 and the coin gives you a heads? Use a list to help you find out.

Solution

Start with a list of all the possible outcomes. 1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T. There are 12 equally-likely outcomes. Of these, only 5H is a five with a heads. Therefore, the answer is 1/12

What you might have noticed is that P(five)= 1/6 and P(heads)=1/2. Curiously, 1/6 * 1/2 = 1/12. Is this just a coincidence or is there something more here? You might recognize that flipping a coin does not affect what you roll on the die. When two events do not have an impact on each other, the events are called independent. Consider the two situations below.

Situation 1: Does a coin have a memory? As far as we can tell, the answer is no. This suggests then that a coin does not pay attention to whether it came up heads or tails. It does not want to make sure that the same number of heads come up as tails. Even if it comes up heads many times in a row, the next flip of that coin is not influenced whatsoever by the previous flips. Successive coin flips are independent of one another.

Situation 2: Suppose your teacher picks students to do problems on the board. After each student does their problem, the teacher gives the student a piece of candy. Because your teacher wants make sure that every student gets a chance to do a problem and get a piece of candy, she keeps track of who has worked problems on the board. The selection of the next student is not independent of previous selections the teacher has made.

[Figure2]

Example 2

Decide which pairs of events below are independent.

i) Two cards are dealt, one after the other, from a standard deck of 52 cards.

ii) A spinner with three colors is spun twice.

iii) A single die is rolled and a coin is flipped.

iv) You play on the school baseball team and you win a carnival game by throwing a baseball to try to break a plate.

Solution

Situations ii) and iii) represent pairs of independent events. The result from the first spin of the spinner does not affect the result of the second spin of the spinner. The result of the roll does not impact what happens when the coin is flipped.

Situations i) and iv) are not independent. Once the first card from the deck is dealt, the probabilities for what the second card might be will change. For example, if the first card was the ace of spades, it is impossible for the second card to also be the ace of spades. Being a baseball player makes it more likely mean that you are accurate and can throw a ball harder than a typical person and you therefore would be more likely to break a plate.

Let's do another example involving calculations and investigate if multiplying probabilities in a situation involving independent events gives us the correct result.

Example 3

A coin is flipped three times in a row. What is the probability that all three flips result in heads? Find your answer by either using a tree diagram or by making a list.

Solution

To be organized, we can make an alphabetically list. HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 8 outcomes altogether and only one of those is HHH. P(HHH)= 1/8

The probability of heads on one flip is 1/2 . As in example 1, we have P(HHH)= 1/2 * 1/2 * 1/2 = 1/8. Note that the three coin flips are independent of each other.

In both Example 1 and Example 3, we could multiply the probabilities of each individual event to get the probability of both events happening. This is always true of independent events. In other words, suppose we want the probability of both some outcome 'A' from one event and some outcome 'B' from a second event that is independent of the first event. If the probability of our first outcome is P(A) and the probability of our second outcome is P(B), then the probability of both A and B happening is P(A and B)= P(A) * P(B)

[Figure3]

Example 4

You are dealt one card from each of two separate decks of cards.

a) What is the probability that both cards are the king of clubs?

b) What is the probability that the two cards are identical?

Solution

In both situations, the events are independent.

a) We have P(K♣ & K♣)=

.

b) There are two good ways to solve this problem. We could imagine that we do what we did in part a) for all 52 cards in the deck. We simply could multiply our answer for part a) by 52 to get

Multiplication of probabilities expands to more than just two independent events. It also works with three or more independent events and it even works with many situations that do not have independent events. In general, when finding the probability of compound events, multiply the probabilities of each individual event. If we are interested in the probability of events A, B, and C happening, we can multiply

.

This same principle also works with compound events in which we distinguish whether or not we have replacement. Suppose we are asked to pick two cards out of a deck. If we are asked to do this without replacement, we will select the first card and record what it is. When we select our second card, we must remember that the deck has changed. Nonetheless, we can find the probability of drawing these two particular cards by multiplying the individual probabilities.

Example 5

Suppose you have a set of pool balls in a bag. You pull two pool balls out of the bag, one after the other, without replacement. What is the probability that both pool balls are striped?

Solution: 7/15

Example 6

Suppose you have a set of pool balls in a bag. You pull two pool balls out of the bag, one after the other, with replacement. (This means that after you record what the first ball is, you put it back into the bag and remix the pool balls before you select the second pool ball.) What is the probability that both pool balls are striped?

[Figure4]

Solution:

There are 7 striped pool balls out of the 15 pool balls. The chance that the first pool ball is striped is 7/15. Since we are going to replace the first pool ball, what is in the bag has not changed. There are still 15 pool balls of which 7 still are striped. Therefore, the chance that the second pool ball is also striped is 7/15. To find the probability that both pool balls are striped, we multiply the individual probabilities to get 49/225. There is approximately a 22% chance that both balls will be striped if we do not use replacement.

We also run into situations where we are dealing with compound events involving very large populations. In these sorts of situations, we must be careful about how we interpret the mathematics.

Example 7

Approximately 20% of all Americans smoke. Suppose two Americans are selected at random. What is the probability that both Americans are smokers?

Solution

The chance that the first person is a smoker is 20%. Some students think that the chance that the second person is a smoker changes after the first person is selected, however, it does not. The population of America is so large that selecting a single person out from that population will not affect the overall percentage of Americans that smoke. The probability that the second person smokes is also 20%. P(2 smokers selected)= 0.2 * 0.2 = 0.04 = 4%

Example 8

Approximately 20% of all Americans smoke. Suppose five Americans are selected at random. What is the probability that all five are non-smokers?

Solution

Since 20% of Americans are smokers, 80% must be non-smokers. This gives us

The chance that all five Americans selected will be non-smokers is about 33%.

BACK