Unlocking Event Probability: Sample Space, Favorable Outcomes, and Experimental Probability

$Favorable outcomes = 15$ , $Total outcomes = 36$

P (Sum \geq 8) = \frac{1 5}{3 6}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

P (Sum \geq 8) = \frac{5}{1 2}

Calculate quotient

P (Sum \geq 8) = 0.416666 \dots

Round to $2$ decimal place(s)

P (Sum \geq 8) \approx 0.42

Consequently, the probability of getting a sum greater than or equal to

8

when two dice are rolled is about

0.42,

42 % .

Notice that the sample space of the experiment contains only

11

outcomes.

{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

However, the outcomes are not equally likely. From the above table, there is greater chance of rolling a

7

than rolling a

2 .

b The experimental probability of an event equals the number of successes obtained divided by the number of trials.

P (Event) = \frac{Number of Successes}{Number of Trials}

Therefore, to find the experimental probability obtained by Ramsha, divide the number of successes she got by the number of trials she conducted. These two numbers can be deduced from the given diagram.

As can be seen, Ramsha got

2

successes in

7

trials. With this information, the experimental probability can be found.

P (Sum \geq 8) = \frac{Number of Successes}{Number of Trials}

Substitute values and simplify

$Number of Successes = 2$ , $Number of Trials = 7$

P (Sum \geq 8) = \frac{2}{7}

Calculate quotient

P (Sum \geq 8) = 0.285714 \dots

Round to $2$ decimal place(s)

P (Sum \geq 8) \approx 0.29

Consequently, for the given event, Ramsha got an experimental probability of

0.29,

29 % .

c Similar to Part B, to find the experimental probability obtained by Mark, divide the number of successes he got by the number of trials he conducted. As before, this information can be found in the given diagram.

Mark got

6

successes in

9

trials. With this information, the experimental probability can be found.

P (Sum \geq 8) = \frac{Number of Successes}{Number of Trials}

Substitute values and simplify

$Number of Successes = 6$ , $Number of Trials = 9$

P (Sum \geq 8) = \frac{6}{9}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

P (Sum \geq 8) = \frac{2}{3}

Calculate quotient

P (Sum \geq 8) = 0.66666 \dots

Round to $2$ decimal place(s)

P (Sum \geq 8) \approx 0.67

Therefore, for the given event, Mark got an experimental probability of

0.67,

67 % .

Discussion

Probability of the Complement of an Event

The probability of drawing a club from a standard deck of cards is $0.25 .$ Knowing this, what is the probability of drawing a spade, heart, or diamond if a card is drawn randomly?

To figure it out, instead of counting the favorable outcomes, the complement rule can be used.

Rule

Complement Rule - Probability

The sum of the probability of an event and the probability of its complement is equal to $1 .$

$P (A) + P (A^{'}) = 1$

This formula is useful when calculating the probability of the complement of an event is easier than calculating the probability of the event itself. Then, the probability of the event is calculated as follows.

P (A) = 1 - P (A^{'})

Proof

Complement Rule

Let

A

be an event,

A^{'}

be its complement, and

S

be the sample space. By the definition of the complement, the union of an event and its complement is equal to the entire sample space.

A \cup A^{'} = S

Because

A \cup A^{'}

and

S

represent the same event, their probabilities are equal.

P (A \cup A^{'}) = P (S)

Since the complement of

A

consists of the outcomes that are not in

A,

events

A

and

A^{'}

are disjoint. By the Addition Rule of Probability, the probability of the union is the sum of the individual probabilities of each of the events.

P (A \cup A^{'}) = P (A) + P (A^{'})

Now the Transitive Property of Equality can be applied to the equalities.

{P (A \cup A^{'}) = P (A) + P (A^{'}) P (A \cup A^{'}) = P (S) ⇓ P (A) + P (A^{'}) = P (S)

Additionally, the probability of the entire sample space

P (S)

is equal to

1 .

By applying the Transitive Property of Equality once more, the rule is proven.

$P (A) + P (A^{'}) = 1$

Using the Subtraction Property of Equality, the formula for the probability of

A

is obtained.

P (A) + P (A^{'}) = 1 ⇕ P (A) = 1 - P (A^{'})

Applying this formula, the probability of drawing a spade, heart, or diamond can be computed.

P (♠, ♡, or ♢) = 1 - P (♣) = 1 - 0.25 = 0.75

Example

Counting Not Multiples of 5 vs. Counting Multiples of 5

Dylan cut out $100$ squares of paper and wrote a number from $1$ to $100$ on each square. He then put the papers in a bag and asked his dad to choose a paper at random.

What is the probability that Dylan's father picks a number that is not a multiple of

5 ?

Hint

What is easier to count, the numbers that are not multiples of $5$ or the numbers that are multiples of $5 ?$ Note that these sets of numbers are the complements of each other.

Solution

Start by remembering what the probability of an event is.

P = \frac{Number of favorable outcomes}{Number of possible outcomes}

Let

A

be the given event, picking a number that is not a multiple of

5 .

However, counting the numbers from

1

100

that are not multiples of

5

can be tedious. In this case, it is worth considering the complement of

A .

Event	Complement
Picking a number that is not a multiple of $5 .$	Picking a number that is a multiple of $5 .$

Counting how many numbers are multiples of

5

is an easier task.

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100

There is a total of

20

favorable outcomes for

A^{'} .

Knowing this, the probability of

A^{'}

can be calculated.

P (A^{'}) = \frac{2 0}{1 0 0} = 0.2

The probability of picking a multiple of

5

0.2 .

Applying the Complement Rule, the probability of

A

can be computed.

P (A) = 1 - P (A^{'})

Substitute

$P (A^{'}) = 0.2$

P (A) = 1 - 0.2

SubTerms

Subtract terms

P (A) = 0.8

Consequently, the probability that Dylan's father picks a number that is not a multiple of

5

0.8 .

Closure

Chances of Picking the Right Door

At the beginning of this lesson, Tearrik was invited by the host of a late-night game show to pick one door out of thee to win a car. However, behind two of the doors are sheep.

a What are the chances that Tearrik will choose the winning door? Round the answer to two decimal places.

b What are the chances of choosing a door with a sheep? Round the answer to two decimal places.

c If there were

70

doors and

20

prizes, what is the chance that Tearrik picks a door with a sheep behind it? Round the answer to two decimal places.

Hint

a Divide the number of favorable outcomes by the number of possible outcomes.

b Use the Complement Rule.

c How many of the doors have sheep behind them?

Solution

a The probability of an event is the quotient of the number of favorable outcomes to the number of possible outcomes.

P = \frac{Number of favorable outcomes}{Number of possible outcomes}

Since there are three doors, there are

3

total possible outcomes, of which only

1

is favorable. Knowing this, the chance that Tearrik will choose the wining door can be calculated.

P (winning door) = \frac{1}{3} \approx 0.33

b Note that the event of picking a door with a sheep is the complement of picking the winning door. Therefore, the Complement Rule can be used.

P (A^{'}) = 1 - P (A)

In Part A, the probability of picking the winning door was found to be about

0.33 .

Substituting this value into the previous equation, the probability of picking a door with a sheep will be obtained.

P (A^{'}) = 1 - 0.33 \approx 0.67

c In this case, there are

70

doors and

20

prizes. This time finding a sheep is the favorable outcome.

Since there are

20

prizes, there are

50

doors with sheep. With this information, the required probability can be found.

P (Sheep) = \frac{Favorable outcomes}{Total outcomes}

$Favorable outcomes = 50$ , $Total outcomes = 70$

P (Sheep) = \frac{5 0}{7 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0}{b / 1 0}$

P (Sheep) = \frac{5}{7}

Calculate quotient

P (Sheep) = 0.71428 \dots