# Simple Probability

### Written by tutor Jeff S.

*Probability* shows us mathematically how likely we are to get a certain *result* for an activity *with predictable outcomes*.

Another way to think of a *result* is to think of the word *outcome*. For example, there are two possible *results* from

flipping a coin: “heads” or “tails”.

Coin flips have *predictable outcomes* because outside factors don’t effect coin tosses. On the other hand, sports events may seem

predictable, but are actually more complex when you think about it. Of course, one team will score more points and win the game and

the other will score fewer points and lose.

However, many factors affect the outcome of the game – making it unpredictable. Will the starting pitcher hurt his arm, or will he

be able to play? If he plays, is he well – rested and ready to play his best today, or did something keep him awake, which will

hurt his performance? These are just a few of the hundreds of things that can affect the outcome of a baseball game.

We need to consider two things when calculating the *probability* of an event: the total number of *outcomes* and the

number of ways we can achieve the desired result. Here’s an example:

Imagine we want to know the probability of getting “heads” when we flip a coin. There are two possible *outcomes*: “heads”

or “tails”. We use this knowledge to write the probability of getting “heads” as a fraction:

** ^{1}/_{2}** * The top number shows how many ways we can arrive at our outcome

(getting “heads” when we

flip a coin).

*The bottom number shows us the total possible number of outcomes.) In this case, we can only get

two outcomes (getting “heads” or “tails”).

One thing to remember is that *probability* doesn’t tell us the exact *outcome* we’ll get. We might get “tails” five times in a row.

Instead, *probability* tells us which outcome we can *reasonably expect*.

Here’s another example. What if our question was, “What is the probability of rolling a 6 on a 6 – sided die?” Since rolling dice has a

*predictable outcome* – we know we’ll roll a number between 1 and 6 – we can use probability. Our answer is:

** ^{1}/_{6}** If we roll one die, there’s only one way to get a six: the six lands face up. We

put the number 1 on

top of the proportion to represent the one way we can get a six if we roll a die. The total possible

number of outcomes is 6 because the die can land on any number from 1 to 6.

## Probability of Dependent Events

*Dependent events* are situations where the *outcome* we get can changes depending on the event that came before it.

For example, suppose we want to know the probability of drawing an ace from a deck of cards. Since the total number of cards in the

deck gets lower each time we draw a card, the probability of drawing an ace will change, too. Here’s how it works:

Since we start with a deck of 52 cards, the probability of drawing an ace is ^{4}/_{52}. There are 4 aces in the deck of 52 cards.

*If* we draw an ace, our probability of drawing another ace is ^{3}/_{51}. There are 3 aces left in the deck of 51 cards.

*If we didn’t draw an ace*, the probability of drawing an ace becomes ^{4}/_{51}. There are still 4 aces left in the deck of 51 cards.

We can use division and multiplication to turn proportions from fractions to decimals and converting them into percentages. Percentages make proportions easier to understand.

For example, if we divide 4 by 51, we get .07. We can multiply this by 100 to turn it into a percentage. When we do this we get 7%. Now we see that there is a 7% chance that we will

draw an ace from the deck of cards if we *don’t draw an ace on our first try*.

This is an example of the *probability of dependent events* – drawing an ace from a deck of cards in our example.

**Exercise 1. Read the sentences below and answer the questions that follow using what you know about dependent probabilities.**

Camila’s 8th birthday is in three weeks and her parents have decided to have a birthday party at their house. They are planning to

have games at the party. They bought small bags of party favors to put into a prize bag. Camila and her friends will reach into the bag

and take out a prize if they come in 1st, 2nd, or 3rd place. There are 15 pairs of sunglasses, 15 animal shaped rubber ink stampers, and

15 small bottles of liquid bubbles in the prize bag.

fraction or a percent in the box below.

Answer: 15/45 or 33%. The bottom number of the proportion represents the total number of outcomes – which is the number of prizes

in the bag. The top number of the proportion represents the number of ways we can achieve the desired result – which is the total

number of pairs of sunglasses in the bag.

as a fraction or a percent in the box below.

Answer: 30/45, or 66%. (30/45 = .66. .66 x 100 = 66%) The bottom number of the proportion represents the total number of

outcomes – which is the number of prizes in the bag. The top number of the proportion represents the number of ways we

can achieve the desired results – which is the total number of ink stampers and liquid bubbles in the prize bag.

## Probability Tree Diagrams

Besides using percentages, we can also use tree diagrams to help us understand the probability of dependent events.

Read the paragraph below to learn how tree diagrams can help us understand

proportions.

Imagine that you recently started working part – time at a movie theater. Two managers, Darrell and Samantha, are the

supervisors on the shifts you work. Both managers occasionally give employees a free movie ticket after their shift. You really

like working with Samantha because she gave you two movie tickets the last 4 times you’ve worked for her. Darrell has only given

you 1 free ticket the last four times he was your supervisor. When you checked the work schedule, you noticed that Samantha works

3 days a week and Darrell works 4. What is the probability that you will get a free movie ticket after your shift if you work 7

days a week?

Tree diagrams help us better understand our chances of getting a free movie ticket. Here’s how to create the diagram:

**Step 1**. Compute the percentage of days each manager works per week and start your diagram. To do this, divide the number of

days each manager works per week by the total number of days in a week (7).

**Step 2**. Compute the probability for each manager’s answer to the question, “Can I have a movie ticket after today’s shift?”

(they can answer “yes” or “no”). To do this, divide the number of times each manager has given each answer by the number of days

you have worked with them (4). Add this information to your diagram. The diagram looks like this:

**Step 3**. Multiply across each row to find the probability for each answer and multiply by 100 to turn it into a percentage.

Here’s what the problems look like: (the final percentages are in bold).

Samantha, “Yes” = .43 x .50 = .215

.215 x 100 = **21.5%**

“No” = .43 x .50 = .215

.215 x 100 = **21.5%**

Darrell, “Yes” = .57 x .25 = .142

.142 x 100 = **14.2%**

“No” = .57 x .75 = .428

.428 x 100 = **42.8%**

When we use a tree diagram, we can see our probabilities much easier. To find the answer to our question,” What is the probability that you will get

a free movie ticket after your shift?”, we add the “yes” percentages together. The problem looks like this:

21.5 + 14.2 = **35.7%**

We can see that we have a 35.7% chance of getting a free movie ticket after our shift if we work part time 7 days a week. Remember that

probability tells us *how likely we are to get a certain result*. It doesn’t tell us exactly what will happen. We might actually get a

free ticket 5 days in a row! Or, we might not get a free ticket for a week. Probability simply helps us *predict* what will happen.

**Exercise 2. Use the tree diagram above to find the answers to the problems.**

is your supervisor the other 2. Samantha has given you a free ticket on 2 of the 3 days. Darrell has given you 1 free ticket the

last 2 days you worked with him. What are the probabilities for your supervisor’s “yes” and “no” answers if asked for a free ticket

now? Convert your answers into percentages and type them in the box below. (Separate your answers with a comma.)

Answer: 39.6, 19.98, 20, 20.

Samantha | Yes | 3/5 = .60 | 2/3 = .66 | .60 x .66 = .396 | .396 x 100 = 39.6% |

Samantha | No | 3/5 = .60 | 1/3 = .333 | .60 x .333 = .1998 | .1998 x 100 = 19.98% |

Darrell | Yes | 2/5 = .40 | 1/2 = .50 | .40 x .50 = .200 | .200 x 100 = 20.0% |

Darrell | No | 2/5 = .40 | 1/2 = .50 | .40 x .50 = .200 | .200 x 100 = 20.0% |

with Samantha 4 days and only 1 with Darrell. Samantha gave you a free ticket 2 of the 4 days and Darrell did not give you a ticket the

last time you worked with him. What are the probabilities for your supervisor’s “yes” and “no” answers now? Convert your answers into

percentages and type them in the box below. (Separate your answers with a comma.)

Answer: 40, 40, 0, 20

Samantha | Yes | 4/5 = .80 | 2/4 = .50 | .80 x .50 = .40 | .40 x 100 = 40% |

Samantha | No | 4/5 = .80 | 2/4 = .50 | .80 x .50 = .40 | .40 x 100 = 40% |

Darrell | Yes | 1/5 = .20 | 0/1 = 0 | .20 x 0 = 0 | 0 x 100 = 0% |

Darrell | No | 1/5 = .40 | 1/1 = 1.0 | .20 x 1 = .20 | .20 x 100 = 20% |