Objective: Students will be able to find the probability of independent and dependent events.

Have you ever been asked to play a game that
involved tossing a coin, rolling a die, or even throwing objects
at a specified target? Did you ever think about whether or not
the game was fair?

In order to determine the fairness of a game we need to examine
some aspects of probability.

**Probability** is the likeliness of some event occurring. For example, what is the probability of flipping
and coin and landing with heads up?

In order to answer this question, we must examine the **sample
space** or the set of all possible outcomes of the toss. In
our coin experiment, the sample space includes only two elements--heads
and tails. The toss itself is called the **event**.

The probability of an event is determined by dividing the number of successes by the total number of outcomes in the sample space. A coin has one (1) head and one (1) tail. If I desire a head on my coin toss and it occurs, that is called a success. There is one head and two possible outcomes in the sample space.

The notation of this probability would be written as P(Heads)= 1/2 or .5.

Above is a picture of a die. To the right are all of the possible outcomes of the roll of one die. Suppose you were asked the probability of rolling a 5. There is only one side of the die that contains a 5 but there are 6 possible outcomes. Therefore the probability of rolling a 5 is 1/6. Mathematically this could be written as P(5)=1/6.

What would be the probability of rolling a prime number?

**Practice problems - Set One:
**Use these problems to determine your
understanding of the material.

1. What is the probability of throwing a six (6) on one roll of a die?

2. Find P(even number) on one roll of a die.

3. Find P(Tails) on one toss of a coin.

4. Using a "funny die" with sides numbered {1,2,3,4,4,5}, Find P(1), P(2), P(3), P(4), P(5).

5. A coin has been tossed 10 times and has come up heads each time. What is the probability that it will land on heads on the eleventh toss?

What is the probability of rolling a seven (7) on a regular six-sided die?

Well since there is no seven on the die, it would be

What is the probability of a coin landing on heads or tails when tossed? Since that

Based on the probability of an impossible event and the probability of a certain event (these being the extremes), the probability of an event can be expressed as a number between 0 and 1.

Referring back to a regular coin, find the sum of the probabilities P(Heads) and P(Tails).

Examine the probabilities for a die.

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) =

1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1

This looks like a pattern. Find the sum of the probabilities of the funny die in the practice problems.

P(1) + P(2) + P(3) + P(4) + P(5) = 1/6 + 1/6 + 1/6 + 2/6 + 1/6 = 6/6 = 1

Sums of probabilities are also used when two events occur that have no elements in common. The sum of the probabilities of

For example, Suppose you roll a standard die. The probability of getting an even number is the sum of the probability of getting a 2, plus the probability of getting a 4, plus the probability of getting a 6.

A card is selected at random from a deck of 52 cards.

1. What is the probability of selecting a black card?

2. What is the probability of selecting a queen?

3. What is the probability of selecting a one (1)?

4. What is the probability of getting a face card?

5. Find the sum P(Hearts) + P(Clubs) + P(Diamonds) + P(Spades) =

It is now time to simulate a game. Given two die (dice), roll the dice. If the dice land on the same number, give yourself a point. If the dice lands on different numbers, give the computer(ME) a point.

Continue this game for approximately two minutes and collect your results on a sheet of paper.

Do you think the game is fair? Before we give the answer to this question, let's look at your results.

Using the results of your simulation, you can write a fraction that represents your results. Using simulations to find the number of times a desired outcome occurs is called experimental probability.

P(Match) = the number of times the dice matched / the total number of rolls.

Below is a list of all of the possible outcomes when the dice are rolled.

1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 3-1, 3-2, 3-3, 3-4, 3-5, 3-6, 4-1, 4-2, 4-3, 4-4, 4-5, 4-6, 5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6.

How many total outcomes are there?

How many are matches?

How many are no matches?

Using the results from our list of all possible outcomes to determine the probabilities is called theoretical probability.

The probability of a match in our game of dice is 6/36.

The probability of no match in our game of dice is 30/36.

The game you just played was not fair. The probability of no match was much greater than the probability of a match. The computer was sure to win.

What makes a game fair? Most of you would agree that a game is fair if each participant has an equal chance to win.

Suppose we had a 12-sided die, called a dodecahedron, whose faces were marked in this fashion: 3, 3, 2, 2, 2, 2, 0, -1, -1, -1, -1, -4. In the game, you are awarded the number of points on the face of the die. Would this game be a game that could be won?

If we set up a table of the possible outcomes and their probabilities, we have

In playing the game, the result of the expected
value determines winnings and losses. A positive expected value
represents wins and a negative expected value represents losses.
The above game gave a positive expected value which means over
a period of time you would have points.

Let's return to our previous question. What makes a game fair? Although we stated that each player should have an equal chance to win, how would that be expressed mathematically?

If two players were playing a game, if one had a positive expected value and the other had a negative expected value, would we think that the game was fair?

In order for the players to have an equal chance against each other or the game itself, the expected value must equal zero (0).

Examine another game that involves money.

A game is played with these rules.

A player wins $3 if he draws an ace, $2 if he draws a king, and $0.50 if he draws a queen or jack.

A player must pay the "kitty" $0.50 if he draws a 10, 9, or 8 and $0.75 if he draws a 7, 6, 5, 4, 3, or 2.

Is it a fair game for the player?

x ----- Ace ----- King ----- Queen/Jack -----10,9,8---7,6,5,4,3,2

P(x) -- 4/52 --- 4/52 ---------8/52 ----------- 12/52-------- 24/52

---------1/13 ----1/13 ---------2/13 ----------- 3/13 --------- 6/13

= 1/13 ( 3 + 2 + 1 + -1.5 + -4.5)

= 1/13( 0) = 0

Because the expected value is 0, the game is fair,
if the player so chooses to play. Does this mean the player will
win? Not necessarily. Actually, if the player plays long enough,
the amount that should be won would be 0. The expected value or
probability may not happen with just a couple of games.

2. Find the expected value if you toss a regular die and subtract one from the square of the number on the face of the die.

3. Suppose you have six coins in your pocket: a penny, a nickel, two dimes, and two quarters. What is the expected value when you pull out one coin?

4. Many people play roulette when visiting Las Vegas. A mechanical wheel contains numbers from 1 to 38. The device randomly selects one of 38 equally likely numbers. If a player chooses one particular number, suppose he gains $35 if his number is selected, but losses $1 if it is not. Calculate the expected value of the game.

5. The 38 numbers in a roulette wheel are colored: 18 red, 18 black and 2 white. If a player chooses a color, he wins a $1 if a number of that color is selected; if not, he losses $1. Is the game fair?

**Go to**** ****Answers
**The above questions were taken from
Introductory Statistics and Probability, A Basis for Decision
Making by David W. Blakeslee and William G. Chinn, Houghton Mifflin
Company, 1975.

If a student has three skirts to wear to school: 1 blue, 1 black, and 1 plaid. She has four blouses: 1 white, 1 red, 1 green and 1 with pink flowers. What is the probability that if she dressed in the dark (choosing her outfit at random), she would wear the plaid skirt with the blouse with pink flowers?

First we will make a tree diagram to view the different outfits possible.

There are 12 possible outfits for the student
to wear. There is only one successful outcome on the tree diagram.
Therefore the P(plaid skirt, pink flower blouse) = 1/12.

There is another way to solve this problem without using a tree diagram. In probabilities involving multiple events where the events are independent of each other use the following:

Use the above rule to answer the following
question.

In a game, a player is to flip a coin and roll a die. What is
the probability of heads on the coin toss and a 6 on the roll
of the die?

Notice the toss of the coin has nothing to
do with the roll of the die. These are independent events.

therefore

P(H,6) = 1/2 * 1/6 = 1/12

If the result of a second event is altered by the event that occurred in the first event, the events are dependent.

The best example of dependent events can be found using a standard deck of cards. Suppose two cards are dealt from a standard deck without replacing the first card drawn. Consider the probabilities of the following:

There are 52 cards at the beginning of the
compound event. Half of the cards are red, so the probability
of a red card on the first draw is 26/52. Once that card is drawn,
regardless of the color, the number of cards to select from has
changed. There are now only 51 cards.

When discussing the probability of a compound
event it is important that we assume the first event is successful.
If the first event is unsuccessful, it would be useless to continue
the compound event.

Therefore in our first draw, we assume a red card
was drawn. In that case, there are only 25 red cards remaining,
so the probability of a red card on the second draw is 25/51.

Find the probability

Again, the draw begins with 4 aces and 52 total
cards.

P(first card is not an ace and second card is
not an ace)=

P(first card is not an ace) * P(second is not an ace/given that
the first card was not an ace)

P(neither card is an ace) = (48/52) * (47/51) = 188/221

2. What is the probability of drawing three cards without replacement so that all of the cards will be face cards?

7 balls are placed in a bag. 4 white, 2 black and 1 gold. Balls are drawn from the bag without replacement.

3. What is the probability that all three balls are white?

4. What is the probability that no ball is white?

5. What is the probability that every ball is a different color?

**Go to Answers**

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