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# Probability

#### Flipping a coin is a well-known probability question. When we flip, it could land on HEADS or TAILS.

• The ACTION is what is happening. In this case, the action is flipping a coin.
• The OUTCOMES are all of the possible results of an action. In this case, there are exactly two outcomes: heads or tails
• An EVENT is any outcome or group of outcomes. In this case, if the coin lands on heads, the event is heads. If we flip the coin twice, and it lands on heads twice, the event is heads and heads.
• When we flip a coin, both outcomes are equally likely to occur - this feature is sometimes called RANDOM.

When we are trying to figure out the PROBABILITY OF AN EVENT (P), we use a ratio to find out how likely it is that an event will happen:

$$Probability(Event)=\frac{number\: of\: favourable\: outcomes}{number\: of\: possible\: outcomes}$$

Example: The probability of a coin landing on tails is 1/2 The number of favorable outcomes (landing tails) is 1, and the number of possible outcomes (landing heads or landing tails) is 2.

P(tails) = 1/2 = 50%

#### Example: What is the probability of the spinner landing on red or purple?

$$Probability(Event)=\frac{number\: of\: favourable\: outcomes}{number\: of\: possible\: outcomes}$$

P (Red or Purple) = 2/5 = 40%

There is a 40% probability that the spinner will land on red or purple.

If a probability question is more complicated, we can make a table to help ourselves.

Example: Bob flips a coin twice. What is the probability that he will flip heads twice?

#### Let's make a table that lists all the possible combinations Bob can have when he flips the coin twice:

Then we plug this info into the probability formula:

$$Probability(Event)=\frac{number\: of\: favourable\: outcomes}{number\: of\: possible\: outcomes}$$

Probability (2 heads) = 1/4 = 25%

In addition to using a chart, we could also draw a tree diagram.

Example: Sue rolls a die twice. What is the probability that she rolls "double sixes" (she rolls 6 twice)?

#### Let's make a tree diagram that lists all of the different possible outcomes:

Then we use the probability formula: $$Probability(Event)=\frac{number\: of\: favourable\: outcomes}{number\: of\: possible\: outcomes}$$

Therefore, out of the 36 possible outcomes, there is a total of 1 outcome that has double sixes.
Probability (double sixes) = 1/36 = 2.8%

#### Compliment of an Event: The compliment of an event is the opposite of the event happening.

The probability of an event plus the probability of its complement always equals 1. In other words, there is a 100% chance that either an event or its complement will happen.

Probability(event) + Probability(complement) = 1

Example: The probability that a student at your school is left-handed is 10%. What is the complement of being left-handed, and what is the probability of the complement?

The complement of being left-handed is right-handed.

If P(left-handed) is 10%, then ...
P(left-handed) + P(right-handed) = 100%
10% + P(right-handed) = 100%
P(right-handed) = 90%

So, the probability that a student at school is right-handed is 90%.

Solved Example 1

The table above shows the distribution of age and gender for 25 people who entered a contest. If the contest winner will be selected at random, what is the probability that the winner will be either a female under age 40 or a male age 40 or older?
A.
4/25
B. 10/25
C. 11/25
D. 16/25

Solution: This is an either-or probability question, which we can tell because the question includes "either" and "or." So we'll be adding two simple probabilities together.

The first probability is the chance of choosing a female under 40 at random from the group of people who entered the contest. There were 8 females under 40 and 25 entrants total according to the table, so our first probability is 8/25.

The second probability is the chance of choosing a male 40 or over. According to the table, there were 2 men 40 or over, and, again, we know that 25 people entered the contest, so our second probability is 2/25.

## The Counting Principle

When there are lots of options, and you want to determine how many different ways you can get something or do something, the COUNTING PRINCIPLE is the way to go.

Let's start with an EXAMPLE: Marissa is getting dressed. She has 4 pairs of pants, 5 shirts and 3 different shoes that could all make a decent outfit. How many different outfits can she make that include 1 pair of pants, 1 shirt, and 1 pair of shoes?

The simplest way to get the answer is to use the counting principle. All you have to do is multiply the number of options for each part of the outfit.

pants x shirts x shoes = total possible combination of outfits
4 x 5 x 3 = 60

A different kind of example: 6 people are running together in a 10K race. How many different ways can they finish the race?

• 6 different people can finish first.
• 1 of the 6 crosses the finish line first.
• Now there are 5 people left for second place.
• Then 4 people are left for 3rd place.
• That leaves 3 people for 4th place.
• Next, there are 2 people in line for 5th place.
• And there is one person left that finishes last.

The counting principle tells us that we can multiply the number of choices for each place to determine how many different ways the race can end. Therefore, the answer is 6 x 5 x 4 x 3 x 2 x 1 = 720.

We also read this is as 6! (6 factorial).
If we had 5! we would multiply 5 x 4 x 3 x 2 x 1.
If we have 4! we would multiply 4 x 3 x 2 x 1.

The ! is a little shortcut so that you do not have to write out all the numbers less than the given number that you need to multiply.

## Independent Events

While playing a board game, you must toss two dice to determine how far you move on the board. The number that you roll on the second die is not affected by the number rolled on the first die. Because one outcome does not affect the other, the events are called INDEPENDENT.

Example: What is the probability of rolling a number less than three on the first die and a number greater than or equal to three on the second die?

Step 1: Determine the probability of each event

A number less than $$3:\frac{2}{6}=\frac{1}{3}$$

A number equal to or greater than $$3:\frac{4}{6}=\frac{2}{3}$$

The combined probability is the multiplication of the probability of the two individual attempts
$$\frac{2}{3}\times \frac{1}{3}=\frac{2}{9}\approx 22%$$

Solved Example 2
You have a fair coin and a fair six-sided die. What is the probability of both getting tails on a coin flip and rolling a 4 on the die?
A.
1/20
B. 1/10
C. 1/12
D. 1/6

## Dependent Events

Imagine that you are one of the captains forming a kickball team. The other captain chooses Lisa, the player that you were about to choose. When it becomes your turn to choose, you have to pick someone else.

Your choice depended on the selection of the other captain. This is an example of a dependent event.

EXAMPLE: Cara has a bag of marbles. There are 15 total. 7 are white, 3 are green, 2 are blue and 3 are purple. What is the probability of choosing a white and then a green, without replacement?

Because the first marble is NOT being replaced, the probability of the second marble depends on the first. When the second marble is chosen, there are only 14 total marbles instead of 15, because one of the marbles has been removed.

Step 1: Determine the probability of the first marble being white: $$\frac{7}{15}$$
Step 2: Determine the probability of the second marble being green: $$\frac{3}{14}$$
Step 3: Multiply the probabilities together to determine the probability of both events occurring.
$$\frac{7}{15}\times \frac{3}{14}=\frac{1}{5}\times \frac{1}{2}=\frac{1}{10}---\rightarrow \frac{1}{10}=0.1=10%$$

There is a 10% chance of grabbing a white and then a green.

Solved Example 3
A hotel offers complimentary breakfast that includes a guest’s choice of one of five types of breakfast cereal, eggs, either bacon, ham, or sausage, and either orange juice or milk. If you order a breakfast combination at random, what are the odds that you will receive a breakfast that includes ham or orange juice?
A. 1/6
B. 1/4
C. 1/3
D. 5/6

Solution: This question isn’t as tough as it looks! You don’t have to take into account anything other than the breakfast meat selections in order to calculate the probability in this case.

While the breakfast choices will differ in the types of cereal they include, we are only interested in picking out how many will include ham and orange juice, so we can simply ignore the information presented about cereal. Of the three choices (bacon, ham, and sausage), we are only interested in one (ham), so that means that 1/3 of the breakfast options will include ham.

Similarly, we’re interested in the half of the choices that include orange juice. We want to find the probability of picking out a breakfast that includes ham OR orange juice, so we add the fractions together:

$$(\frac{1}{3}\cdot \frac{2}{2})+(\frac{1}{2}\cdot \frac{3}{3})=\frac{2}{6}+\frac{3}{6}=\frac{5}{6}$$

$$\frac{5}{6}$$ of the breakfast choices include both ham and orange juice. We can check our answer by considering that these choices will differ in only one aspect: the type of cereal included with each. Since there are five types of cereal offered, it stands to reason that there should be five options that include both ham and orange juice. This is true, as $$\frac{1}{6}\times 30=5$$

Solved Example 4
At your local no-kill animal shelter, 1/3 of the pets available for adoption are cats, and of those cats, 2/5 are kittens; the rest are adult cats. Half of the cats at the shelter have green eyes, and a third of them are declawed. If someone adopts a pet at random from this shelter, what are the odds that they will adopt an adult cat with green eyes that has not been declawed?
A. 1/45
B. 1/30
C. 1/25
D. 1/20

Solution: Let’s write out all of the information we’ve been given:
$$P_{cat}=\frac{1}{3}$$
$$P_{kitten}=\frac{2}{5}$$
$$P_{cat\: with\: green\: eyes}=P_{cat}\cdot \frac{2}{5}$$
$$P_{declawed\: cat}=P_{cat}\cdot \frac{1}{3}$$

That’s a lot of information! Next, let’s construct an expression of the probability we need to calculate: we’re looking for the odds of someone adopting at random an adult declawed cat with green eyes.

$$P_{declawed\: adult\: cat\: with\: green\: eyes }= \frac{1}{3}\cdot (1-\frac{1}{3})\cdot \frac{1}{3}\cdot \frac{1}{2}=\frac{1}{3}\cdot \frac{2}{5}\cdot \frac{1}{3}\cdot \frac{1}{2}=\frac{2}{60}=\frac{1}{30}$$

## Conditional Probability

Very occasionally, the test will hit you with a simple conditional probability question.

Conditional probability is the chances of an event (B) happening given that another event or condition (A) has already happened or been fulfilled. It's still simple probability - desired outcomes over total outcomes - but figuring out the correct number of desired vs. total outcomes can be a little tricky.

Example: There are 100 people working on a performance: 52 dancers, 12 stage technicians, and 36 musicians. Among the dancers, 14 are ballet dancers, 20 are jazz dancers, and 18 are modern dancers. What is the probability of selecting a ballet dancer from those working on the performance, given that the person selected is a dancer?

It might seem like this is asking you the probability of selecting a ballet dancer (of which there are 14) from everyone working on the performance (of which there are 100).

But actually, it's asking you the probability of selecting a ballet dancer from the dancers, because we are accepting as a given (as a condition) that the person we are randomly selecting is a dancer. We can tell this from the phrase "given that the person selected is a dancer."

Thus, we must calculate the probability of selecting a ballerina (Event B) given condition A, that the person we select will be from among the 52 dancers. So the answer is 14/52

#### Solved Example 5

The same 20 contestants, on each of 3 days, answered 5 questions in order to win a prize. Each contestant received 1 point for each correct answer. The number of contestants receiving a given score on each day is shown in the table above.

No contestant received the same score on two different days. If a contestant is selected at random, what is the probability that the selected contestant received a score of 5 on Day 2 or Day 3, given that the contestant received a score of 5 on one of the 3 days?
A. 1/7
B. 3/7
C. 5/7
D. 6/7

Solution: This grid-in question is both an either/or probability question and a conditional probability question. The condition is we are selecting at random among contestants who received a 5/5 on one of the three days. The question tells us that we can assume that no one got the same score multiple days, so we can assume that the 7 people who received 5s across the 3 days are all different people.

Our total number of potential outcomes, then, is 7 - not 20, the total number of contestants.

Our desired outcome is the number of people receiving 5s on Day 2 or Day 3 (there's that either/or)! 2 people received 5s on Day 2, and 3 on Day 3. The two probabilities that we are adding together, then, are

$$\frac{2}{7}+\frac{3}{7}=\frac{5}{7}$$