Download C1 Theoretical Probability Probability measures the chance

C1 Theoretical Probability
Probability measures the chance that an uncertain event will occur. We use ratios to describe
how likely or unlikely an outcome will be. For example, the ratio 3/6 = ½ represents the
probability of rolling an even number on a six-sided dice.
Mathematically, the theoretical probability of an event is the number of ways that the event can
occur, divided by the total number of outcomes. Written as a ratio, we get
P(event) =
number of favorable outcomes
number of possible outcomes
There are four major terms that are associated with probability.
• An outcome is the result of an experiment. In the experiment of rolling a single die, one
outcome was rolling a 2.
• The sample space, S, is the set of all possible outcomes often written as a list with
brackets. In the experiment of rolling a single die, S = {1, 2, 3, 4, 5, 6}.
• An event, E, is a list of outcomes that are a subset of the sample space. In the experiment
of rolling a single die, E is the event of rolling an even number or E = { 2, 4, 6}
When each outcome is equally likely, we can redefine the theoretical probability of an event, E,
number of elements in E
as P(E) =
number of elements in S
Given a single die with faces numbered 1 – 6, the probability of tossing an even number on one
throw of the dice is 3/6 or ½ since there are 3 favorable outcomes out of 6 possible outcomes.
As the chart below shows, probabilities range from 0 to 1. If an event is impossible and will
never occur, the probability is 0. If an event is absolutely certain to occur, the probability is 1.
Otherwise, the value of a probability is between 0 and 1.
C2 Sample Space and Tree Diagrams
If all outcomes are equally likely, we can find the probability of an event by listing possible
outcomes. A sample space is a set of all possible outcomes for an activity or experiment. As
you can see from the following table, sample spaces can become very large.
Activity
Rolling a die
Rolling a die
Tossing a coin 3
times
Sample Space
There will be 6 outcomes in
the sample space:
S = {1, 2, 3, 4, 5, 6}
There will be 6 outcomes in
the sample space:
S = {1, 2, 3, 4, 5, 6}
Probability
What is the probability of
rolling a 3 or 5?
E = {3, 5} P(E) = 2/6 = 1/3
P(3 or 5) =P(3)+P(5)= 1/6 + 1/6
What is the probability of
rolling an odd or prime
number?
E = {1, 2, 3, 5} P(E) = 4/6 =
2/3
What is the probability of
There will be 8 outcomes in flipping at least two heads?
the sample space:
S = {HHH, HHT, HTH, THH, E = {HHT, HTH, THH, HHH}
HTT, THT, TTH, HHH}
P(E) = 4/8
There will be 52 cards in the
Drawing a card from sample space:
S = {Spades: 2,3,4,5,6,7,8,9,10, ace,
a standard deck
jack queen, king,
Clubs: 2,3,4,5,6,7,8,9,10, ace, jack,
queen, king,
Diamonds: 2,3,4,5,6,7,8,9,10, ace, jack,
queen, king,
Hearts: 2,3,4,5,6,7,8,9,10, ace, jack,
queen, king}
What is the probability of
selecting a jack from this deck?
P(E) = 4/52 = 1/13
What is the probability of
Drawing one marble
from the bottle
There will be 8 marbles in the selecting a marble that is NOT
blue from this bottle?
sample space:
S = {blue marble, blue marble,
blue marble, blue marble, blue P(NOT Blue) = P(Red) =3/8
marble, red marble, red marble,
red marble}
P(NOT Blue) = 1 – P(Blue) =
1 – 5/8 = 3/8
There will be 36 outcomes in What is the probability that the
the sample space:
dice mathch?
Rolling a pair of dice S = {(1,1) (1,2) (1,3) (1,4) (1,5)
(1,6)
E={(1,1), (2,2), (3,3), (4,4), (5,5),
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (6,6)}
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) P(E) = 6/36 = 3/36
(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)
(6,6)}
Choosing an outfit
from a green blouse,
a red blouse, a
yellow blouse, a
black skirt, and a
pair of jeans
There will be 6 outfit
combinations in the sample
space:
What is the probability that
your outfit will NOT have a
yellow blouse?
{green blouse-skirt, green
blouse-jeans, red blouse-skirt, P(E) = 4/6 = 2/3
red blouse-jeans, yellow blouseskirt, yellow blouse-jeans}
Mutually Exclusive Events: In the first experiment, rolling a die, the event “getting a 3” and the
event “getting a 5” is an example of mutually exclusive events. Two events are mutually
exclusive or disjoint if they share no common outcomes, that is, the events can not happen at the
same time. In this experiment, we can not roll both a 3 and a 5 at the same time. This idea
allows us to find the P(3 OR 5) in two ways. By counting the number of favorable outcomes or
finding the sum of the individual probabilities of each event as shown above. This is the
foundation for an important property of probability.
The Addition Property of Mutually Exclusive Events
If events E and F are mutually exclusive, then P(E or F) = P(E) + P(F)
On the other hand, the second experiment from the table above gives us an example of two
events that are not mutually exclusive. The event “getting an odd number” and the event
“getting prime number’ can happen at the same time since there are two odd prime numbers in
the sample space, 3 and 5. In this example, we can not use the Addition Property to find the
probability of “getting an odd” OR “getting a prime” as noted below.
P(Odd OR Prime) = 4/6 BUT P(Odd – 1,3,5) + P(Prime- 2,3,5) = 3/6 + 3/6 = 1
Since the events of rolling a prime and rolling an even have two elements in common, we need
to eliminate the chance of double counting them in our addition formula. If we eliminated the
probability of getting the common event, P(Odd And Prime) = 2/6, we would get the correct
probability. That is,
P(Odd OR Prime) = P(Odd) + P(Prime) – P(Odd AND Prime) = 3/6 + 3/6 – 2/6 = 4/6.
Therefore, in the case where the events are not mutually exclusive, we need to modify the
addition formula as shown below.
The Addition Property for Two Events
If E and F are two events, then P(E or F) = P(E) + P(F) – P(E and F)
Notice that the above formula works even for mutually exclusive events. In the case P(E AND F)
= 0.
Complimentary Events: In the 5th experiment, drawing a marble, consisted of two possible
outcomes, “selecting a Red marble” or “selecting a Blue marble”. Since one of these two events
must occur, we call these events compliments. That is, the event “NOT Blue” is the compliment
of the event of selecting a Blue marble from the bottle. Since one of the two events (drawing a
blue marble and not drawing a blue marble) must happen, the P(Blue) + P(NOT Blue) = 1.
Complementary events, therefore, share a simple property of probability: P(NOT Blue) = 1 –
P(Blue).
Probabilities of Complementary Events
If events E and G are complementary events, then P(E) = 1 – P(G).
When determining a sample space, we must make sure that we consider ALL possibilities. As
the sample space gets larger, it can be useful to construct tree diagram or use the Counting
Principle.
A tree diagram is a useful model for finding the sample
a two-part (or more) experiment such as flipping a coin
rolling a die. In this example there are two different
outcomes in the first stage as the coin can land heads up
up and 6 different outcomes for the second stage of
the die. By following the different paths in the tree
we can determine that there are 12 possible outcomes in
sample space. Sample space: S = { H1, H2, H3, H4,
T1, T2, T3, T4, T5, T6 }
space of
and
or tails
rolling
diagram,
the
H5, H6,
Notice, that in the tree diagram each of the two possible
outcomes in the first stage (H, T) is associated with one
of six
different outcomes in the second stge. So in total, there must be 2 x 6 = 12 outcomes in the
sample space. This important property is called the Fundamental Counting Principle. This
principle can be used as an efficient method to count the total number of outcomes in two- or
more-part experiments.
The Fundamental Counting Principle
If an event E has e outcomes and event F has f outcomes, then the experiment that has
event E followed by event F has e x f outcomes.
The Fundamental Counting Principle can be generalized to more than two events as detailed in
the Counting Problems Learning Object
Remember, finding the probability often requires you to be able to count all the possible
outcomes on an experiment, as well as all the favorable outcomes. In the two-part experiment of
flipping coin and rolling a die, we can see that the probability of each event, such as Heads AND
a 1 (H1), is 1/12 since there is one favorable outcome out of 12. The probability of a Heads
followed by an Even number is given by counting the branches of the tree diagram that lead to
this favorable event: H2, H4, and H6. This means P(H and E) = 3/12 or ¼. That is, we would
expect that 25% of the time we would get this favorable result.
Independent and Dependent Events: If the weather forecast is cloudy, it affects the probability
that you will bring an umbrella to work. The events “cloudy forecast” and “bringing an
umbrella” are referred to as dependent events since the probability of one event occurring
depends on the other event.
Two events E and F are independent if the probability of
first event is not affected by the results of the second event.
example, in the coin-die experiment, the probability that an
number is rolled of the die is not affected by the event that
coin landed on Heads. That is, the outcome of the die is not
altered by the outcome of the coin as these two events are
separate events.
the
For
Even
the
In the section above, we calculated the probability of Heads
1 through the tree diagram. From the diagram, we see that
probability of flipping a heads is ½ followed by the
probability of rolling a 1 is 1/6. Notice that the probability
H1 is ½ x 1/6 = 1/12. This example gives rise to the
independent events probability formula.
AND
the
of
Independent Events and Probability
If E and F are independent events, then P(E and F) = P(E) x P(F).
To find the probability of dependent events, we will use the example of selecting
two marbles from the bottle to the left. Suppose we select two marbles WITHOUT
REPLACEMENT. This means that we select one marble and draw a second marble
without putting the first marble back into the bottle. These events are dependent
events because the probability of selecting a blue marble on the second draw is
dependent on whether the first marble was blue or red.
Suppose we want to find the probability that both marbles are red. The probability that the first
marble is red is given by P(Red) = 3/8. If we do not replace this red marble, then there are only
two red marbles and seven total marbles. The probability that the second marble is red given the
first marble was red is given by P(Red given Red) = 2/7. To find the probability that these two
dependent events occur in succession we can use a multiplication rule. The probability of Red
on the 1st draw and Red on the 2nd draw is given by
P(Red 1st and Red 2nd) = P(Red 1st ) x P(Red 2nd given Red 1st) = 3/8 x 2/7 = 2/56 = 1/28
Multiplication Rule for Probability
If E and F are two events, then P(E and F) = P(E) x P(F given E).
Notice that in this probability rule, the events E and F may or may not be independent. If the
events are independent that knowing that event E has occurred does not alter the probability of
event F. In the coin-die experiment if E is the event that the coin lands on Heads and F is the
event that the die lands on 1, then P(F given E) = P(1 given H) = P(1) = 1/6. In other words, if
the events E and F are independent the P(F given E) = P(F).
To practice these probability rules please visit the following learning objects.
C3 Expected Value
State lotteries and casinos use probability to determine that how much they would expect to pay
out in the long run. This expected amount is referred to as the expected value.
Consider the following game in which you roll two die. If the dice match, you win $3. If the die
do not match you must pay your opponent $1. If you played this game over and over, how much
would you expect to win/lose? To find this value we must first determine the probabilities
associated with the complimentary events: the die match and the dice do not match. From the
probability table above, we see that P(dice match) = 6/36 or 1/6. The probability of the
complement is given by P(not match) = 1 – 1/6 = 5/6. The expected value is given by the
following sum.
$2(Probability of Winning) + (-$1)(Probability of Losing) =
2 x P(Dice Match) + (-1) x P(Not Match) =
2 x (1/6) + (-1) x (5/6) =
2/6 + (-5/6) =
-3/6 = -1/2 = -$0.50.
This expected value means that on average you would expect to lose $0.50