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Independent and Dependent
Events
Topic Index | Algebra Index | Regents Exam Prep Center
Two events are said to be independent if the result of the second event is not affected by the
result of the first event.
If A and B are independent events, the probability of both events occurring is the product of the
probabilities of the individual events
If A and B are independent events,
P(A and B) = P(A) • P(B).
Example:
A drawer contains 3 red paperclips, 4 green paperclips, and 5 blue paperclips.
One paperclip is taken from the drawer and then replaced. Another paperclip is taken from the
drawer. What is the probability that the first paperclip is red and the second paperclip is blue?
Because the first paper clip is replaced, the sample space of 12 paperclips does not change from the
first event to the second event. The events are independent.
P(red then blue) = P(red) • P(blue) = 3/12 • 5/12 = 15/144 = 5/48.
If the result of one event IS affected by the result of another event, the events are said to be
dependent.
If A and B are dependent events, the probability of both events occurring is the product of the
probability of the first event and the probability of the second event once the first event has
occurred.
If A and B are dependent events,
and A occurs first,
P(A and B) = P(A) • P(B,once A has occurred)
... and is written as ...
P(A and B) = P(A) • P(B|A)
Example:
A drawer contains 3 red paperclips, 4 green paperclips, and 5 blue paperclips.
One paperclip is taken from the drawer and is NOT replaced. Another paperclip is taken from
the drawer. What is the probability that the first paperclip is red and the second paperclip is
blue?
Because the first paper clip is NOT replaced, the sample space of the second event is changed. The
sample space of the first event is 12 paperclips, but the sample space of the second event is now 11
paperclips. The events are dependent.
P(red then blue) = P(red) • P(blue) = 3/12 • 5/11 = 15/132 = 5/44.
Don't panic! You can do it!
Probability is mostly common
sense.
Intuitive Idea of Probability
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We use ratios to show how likely, or unlikely, an outcome might be.
This ratio is called the probability.
For example, the probability of getting heads on the toss of a penny is the
ratio
. Heads is one of two possible outcomes when a penny is tossed.
Our ratio for probability can be thought of as:
Ratios are most often expressed in the form of a fraction. Fractions, however, can also be expressed as
decimals or percents. So probabilities may be expressed as fractions, decimals or percents.
As the chart on the left 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.
Events that are likely to occur have a probability greater than 0.5.
Events that are not likely to occur have a probability less than 0.5.
While probability was originally developed to deal with gaming techniques, it is used today in a
wide range of applications. Even common sayings reflect the concept of probability:
Someone may say "Don't hold your breath" when you are waiting for
something to happen. What is the person implying about the
probability that the event will occur in the near future?
ANSWER: This statement implies that you may die holding your breath waiting for this event to occur.
Obviously, the probability of the event occurring in the near future, is extremely low, if not impossible.
Example:
A fair die is rolled, and the outcome noted. Determine whether each of the following outcomes
is: certain to happen, certain not to happen, likely to happen, likely to not happen.
•
•
•
•
•
•
•
•
rolling a 2
rolling a number less than 3
rolling a 7
rolling a number less than 10
rolling a number greater than 1
rolling a factor or 6 (1,2,3,6)
rolling a number that is an integer
rolling a negative number
•
•
•
•
•
•
•
•
probability = 1/6 likely to not happen
probability = 1/6 + 1/6 = 2/6 likely not to happen
not possible - certain not to happen
certain to happen
probability = 5/6 likely to happen
probability = 4/6 likely to happen
certain to happen
certain not to happen
Theoretical (experimental) vs
Empirical Probability
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Probability
describes the chance that an
uncertain event will occur.
Empirical (experimental) Probability of an event is an "estimate" that the event
will happen based on how often the event occurs after collecting data or running an experiment (in
a large number of trials). It is based specifically on direct observations or experiences.
Empirical (experimental) Probability
Formula
P(E) = probability that an event, E, will occur.
top = number of ways the specific event occurs.
bottom = number of ways the experiment could
occur.
Example: A survey was conducted to
determine students' favorite breeds of dogs. Each
student chose only one breed.
Dog Collie Spaniel Lab Boxer PitBull Other
10
15
35
8
5
12
#
What is the probability that a student's favorite
dog breed is Lab?
Answer: 35 out of the 85 students chose Lab.
The probability is
.
Theoretical Probability
of an event is the number of ways that the event can occur,
divided by the total number of outcomes. It is finding the probability of events that come from a
sample space of known equally likely outcomes.
Theoretical Probability Formula
P(E) = probability that an event, E, will occur.
n(E) = number of equally likely outcomes of E.
n(S) = number of equally likely outcomes of sample
space S.
Example 1: Find the probability of rolling a
six on a fair die.
Answer: The sample space for rolling is die is 6
equally likely results: {1, 2, 3, 4, 5, 6}.
The probability of rolling a 6 is one out of 6 or
.
Example 2: Find the probability of tossing a fair die and getting an odd number.
Answer:
event E : tossing an odd number
outcomes in E: {1, 3, 5}
sample space S: {1, 2, 3, 4, 5, 6}
Comparing Empirical (experimental) and Theoretical Probabilities:
Karen and Jason roll two dice 50 times and record their
results in the accompanying chart.
1.) What is their empirical probability of rolling a 7?
2.) What is the theoretical probability of rolling a 7?
3.) How do the empirical and theoretical probabilities
compare?
Sum of the rolls of two dice
3, 5, 5, 4, 6, 7, 7, 5, 9, 10,
12, 9, 6, 5, 7, 8, 7, 4, 11, 6,
8, 8, 10, 6, 7, 4, 4, 5, 7, 9,
9, 7, 8, 11, 6, 5, 4, 7, 7, 4,
3, 6, 7, 7, 7, 8, 6, 7, 8, 9
Solution:
1.) Empirical probability (experimental probability or observed
probability) is 13/50 = 26%.
2.) Theoretical probability (based upon what is possible when
working with two dice) = 6/36 = 1/6 = 16.7% (check out the table
at the right of possible sums when rolling two dice).
3.) Karen and Jason rolled more 7's than would be expected
theoretically.
Vocabulary to keep in mind:
Biased: one result has a better chance of happening
than another result.
Unbiased (fair) : each result has an equal chance of happening.
Mutually Exclusive Events
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An event is a set of outcomes. It is a subset of the sample space for an activity or experiment.
Event: Drawing a black card from a deck of standard cards.
Probability of this event = 26/52 = 1/2
When an event corresponds to a single outcome of the activity, it is often called a simple event.
Simple Event: Drawing the queen of spades from a deck of standard cards.
Probability of this event = 1/52
Two events that have NO outcomes in common are called mutually exclusive.
These are events that cannot occur at the same time.
Mutually exclusive - think of this as the 2 events
together (mutually) agreeing to exclude (not
include) each others' elements.
They have agreed to be different - mutually
exclusive.
Examples:
A pair of dice is rolled. The events of rolling a 6 and of rolling a double have the outcome (3,3)
in common. These two events are NOT mutually exclusive.
A pair of dice is rolled. The events of rolling a 9 and of rolling a double have NO outcomes in
common. These two events ARE mutually exclusive.
For any two mutually exclusive events, the probability that an outcome will be in one event or
the other event is the sum of their individual probabilities.
If A and B are mutually exclusive events,
P(A or B) = P(A) + P(B)
For any two events which are not mutually exclusive, the probability that an outcome will be in
one event or the other event is the sum of their individual probabilities minus the probability of
the outcome being in both events.
Look out!! Don't get stuck on this one!!!
If events A and B are NOT mutually exclusive,
P(A or B) = P(A) + P(B) - P(A and B)
Example 1: A pair of dice is rolled.
What is the probability that the sum of the numbers
rolled is either 7 or 11?
Six outcomes have a sum of 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
P(7) = 6/36
Two outcomes have a sum of 11:
(5,6), (6,5)
P(11) = 2/36
The sum of the numbers cannot be 7 and 11 at the same time, so these events are mutually exclusive.
P(7 or 11) = P(7) + P(11) = 6/36 + 2/36 = 8/36 = 2/9
Example 2: A pair of dice is rolled.
What is the probability that the sum of the numbers
rolled is either an even number or a multiple of 3?
Of the 36 possible outcomes, 18 are even sums.
P(even) = 18/36 = 1/2
Sums of 3, 6, 9, and 12 are multiples of 3.
There are 12 sums that are multiples of 3.
P(multiple of 3)= 12/36 = 1/3
However, some of these outcomes appear in both events.
(not mutually exclusive)
The sums that are even and a multiple of 3 are 6 and 12.
There are 6 ordered pairs with these sums.
P(even AND a multiple of 3) = 6/36 = 1/6
P(even OR a multiple of 3) = 18/36 + 12/36 - 6/36 = 24/36 = 2/3
Conditional Probability
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The conditional probability of an event B, in relation to event A, is the probability
that event B will occur given the knowledge that an event A has already occurred.
In plain English ...
You toss two pennies. The first penny shows HEADS and the other penny
rolls under the table and you cannot see it. Now, what is the probability
that they are both HEADS? Since you already know that one is HEADS,
the probability of getting HEADS on the second penny is 1 out of 2.
The probability changes if you have partial information.
This "affected" probability is called conditional probability.
Notation for conditional probability: P(B|A)
read ... the probability of B given A.
To establish our formulas for conditional probability, we will need to revisit our previous
discussion of independent and dependent events.
• If events A and B are independent (where event A has no effect on the probability of event
B), then the conditional probability of event B given event A is simply the probability of
event B.
Example: Two colored dice (one blue, one yellow) are rolled.
a. What is the probability of rolling "box cars" (two sixes)?
b. What is the probability of rolling "box cars" knowing the first toss is a six?
Answer:
a. The probability of getting "box cars" (two sixes) is
.
b. If, however, we roll the dice and and see that the blue die shows a six (and the yellow
die is out of sight), the probability of the yellow die being six is
.
The probability of rolling "box cars", knowing that the first roll is a six, is
.
The probability changes when you have partial information about the situation.
• If events A and B are dependent (where event A has effect on the probability of event B),
then we saw that the probability that both events occur is defined by:
P(A and B) = P(A) • P(B|A).
Dividing both sides of this equation by P(A) gives us our formula for conditional probability of
event B given event A, where event A affects the probability of event B:
Assuming P(A), n(A) are not zero.
Example 1: A bag contains 12 red M&Ms, 12 blue M&Ms, and 12 green M&Ms.
What is the probability of drawing two M&Ms of the same color in a row?
Answer:
Intuitive: There are a total of 36 M&Ms in the bag. You draw a blue M&M and eat it.
There are now 11 blue M&Ms remaining in the bag. There are 35 total M&Ms now
remaining. You will now need to draw another blue M&M. The conditional probability
will be:
P(Draw blue M&M | First M&M was blue) =
.
Using the formula:
P(A and B) =
P(A) =
P(Draw same as first color M&M | First M&M color) = P(B | A) =
Example 2: In a school of 1200 students, 250 are seniors, 150 students take math, and
40 students are seniors and are also taking math. What is the probability that a randomly
chosen student who is a senior, is taking math?
Answer: These questions can be confusing. It sounds, at first read, that they are asking
for the probability of choosing a student who is a senior and who is taking math. Not
quite right!
It helps to re-word the question into:
Find the probability that the student is taking math, given that the student is a senior.
B = the student is taking math
n(A) = the student is a senior = 250.
n(A and B) = the student is a senior and is taking math = 40.
Probability
Involving AND and OR
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Let's examine AND first:
In probability, an
outcome is in event
"A and B" only when
the outcome is in both
event A and event B.
In Venn Diagrams, we learned that an
element was in the intersection "A and B",
only when the element was in BOTH sets.
Rule (for AND):
n(A and B) means the number of outcomes in both A and B.
n(S) means the total number of possible outcomes
Example:
A die is rolled. What is the probability that the number is even and less than 4?
Event A: Numbers on a die that are even: 2, 4, 6
Event B: Numbers on a die that are less than 4: 1, 2, 3
There is only one number (2) that is in both events A and B.
Total outcomes S: Numbers on a die: 1, 2, 3, 4, 5, 6 (total = 6)
Answer:
Probability = 1/6
Let's examine OR:
In probability, an
outcome is in event
"A or B" when the
outcome is in either
(or both) event A or
event B.
In Venn Diagrams, we learned that an
element was in the union "A or B",
when the element was in either or both
sets.
Rule (for OR):
The rule for OR takes into account those values that may get counted more than once when the
probability is determined. Check out the example below.
Example:
A die is rolled. What is the probability that the number is even or less than 4?
Event A: Numbers on a die that are even: 2, 4, 6
P(A)=3/6
Event B: Numbers on a die that are less than 4: 1, 2, 3 P(B)=3/6
P(A and B) = 1/6 (see rule above)
Answer: Probability = P(A) + P(B) - P(A and B)
= 3/6 + 3/6 - 1/6 = 5/6
**Notice in this problem that the number 2 appears in both event A and event
B. If we did not subtract the P(A and B), the answer would be 1 - which we
know is not true since the number 5 appears in neither event.
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