The Multiplication Rules and Conditional Probability
The Multiplication Rules and Conditional Probability
Multiplication Rules:
Objective: Find the probability of compound events using multiplication rules.
The multiplication rules can be used to find the probability of two or more events that occur in sequence. For example, if a coin is tossed then a die is rolled, one can find the probability of getting a head on the coin and a 4 on the die. These two events are said to be independent since the outcome of the first event (tossing a coin) does not affect the probability outcome of the second event (rolling a die)
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Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring
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Here are other examples of independent events:
Rolling a die and getting a 6, and then rolling a second die and getting a 3.
Drawing a card from a deck and getting a queen, replacing it and drawing a second card and getting a queen
For example, if a coin is tossed twice, the probability of getting two heads is
1/2 • 1/2 = 1/4
This result can be verified by looking at the sample space, HH, HT, TH, TT
Multiplication Rule 1:
When two events are independent, the probability of both occurring is
P(A and B) = P(A)•P(B)
Example 1:
A is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die.
Solution:
P(head and 4) = P(head) P(4) = 1/2 • 1/6 = 1/12
Note that the sample space for the coin is H, T; and for the die it is 1, 2, 3, 4, 5, 6.
The problem given in above example can also be solved by using the sample space:
H1 H2 H3 H4 H5 H6 TI T2 T3 T4 T5 T6
The solution is 1/2, since there is only one way to get the head-4 outcome.
Example 2:
A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace.
Solution:
The probability of getting a queen is 4/52, and since the card is replaced,
the probability of getting an ace is 4/52. Hence, the probability of getting a queen and an ace is
P(queen and ace) = P(queen) • P(ace)
= 4/52 • 4/52
= 16/2704
= 1/169
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Example: An urn contains three red balls, two blue balls, and five white balls. A ball is selected and its color noted.Then it is replaced. A second ball is selected and its color noted. Find the probability of each of the following.
a. Selecting two blue balls
b. Selecting a blue ball and then a white ball
c. Selecting a red ball and then blue ball
Solution:
a. P(blue and blue) = P(blue) * P(blue)
= 2/10 * 2/10
= 4/100
= 1/25
b. P(blue and white) = P(blue) * P(white)
= 2/10 * 5/10
= 10/100
= 1/10
c. P(red and blue) = P(red) * P{blue)
= 3/10 * 2/10
= 6/100
= 3/50
Multiplication rule 1 using can be extended to three or more independent events by the formula
P(A and B and C and....... and K) = P(A) P(B) P(C) .......... P(K)
Small Sample and Large Population:
When a small sample is selected from a large population and the subjects are not replaced, the probability of the event occuring changes so slightly that for the most part, it is considered to remain the same. The next two examples illustrate this concept.
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Example 1:
A Harris poll found that 46% of Americans say they suffer great stress at least once a
week. If three people are selected at random, find the probability that all three will say
that they suffer great stress at least once a week.
Solution:
Let S denote stress. Then
P(S and S and S) = P(S) • P(S) • P(S)
= (0.46) (0.46) (0.46)
≅ 0.097
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Example 2:
Approximately 9% of men have a type of color blindness that prevents them from distinguishing between red and green. If 3 men are selected at random, find the probability that all of them will have this type of red-green color blindness.
Solution:
Let C denote red-green color blindness. Then,
P(C and C and C) = P(C)•P(C)•P(C)
= (0.09) (0.09) (0.09)
= 0.000729
Hence, the rounded probability is 0.0007.
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In the above two examples, the events were independent of each other, since occurrence of the first event in no way affected the outcome of the second event. On the other hand, when the occurrence of the first event changes the probability of the occurance of the second event, the two events are said to be dependent.
For example, suppose a card is drawn from a pack and not replaced, and then a second card is drawn. what is the probability of selecting an ace on the first card and a king on the second card'?
Before an answer to the question can be given, one must realize that the events are dependent.
The probability of selecting an ace on the fist draw is 4/52,(because there are 4 aces and 52 total cards)
if that card is not replaced, probability of selecting a king on the second card is 4/51, Since there are four kings and 51 cards remaining.
The outcome of the first draw has affected the outcome of the second draw.
When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way probability is changed the events are said to be dependent.
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Here are some examples of dependent events:
Drawing a card from a deck, not replacing it, and then drawing a second card
Selecting a ball from an urn, not replacing it, and then selecting a second ball
Having high grades and getting a scholarship
Parking in a no-parking zone and getting a parking ticket
In order to find probabilities when events are dependent, use the multiplication rule with a modification in notation. For the problem just discussed, the probability of getting an ace on the first draw is 4/52,and the probability getting a king on the second draw is 4/51.
By the multiplication rule, the probability of both events occurring is
4/52 • 4/51
= 16/2652
= 4/663
The event of getting a king on the second draw given that an ace was drawn first time is called a conditional probability.
The conditional probability of an event B in relationship to an event A is the probability that notation for event B occurs after event A has already occurred. The notation for conditional probability is P(B|A). This notation does not mean that B is divided by A;
rather, it means the probability that event B occurs given that event A has already occurred. In the card example, P(B\A) is the probability that the second card is a king
given that the first card is card an ace, and it is equal to 4/51, since the first card was not
replaced.
Multiplication Rule 2:
When two events are dependent, the probability of both occurring is
P(A and B) = P(A) P(B|A)
Example 1:
In a shipment of 25 microwave ovens, 2 are defective. If two ovens are
selected and tested, find the probability that both are defective if the first one is not re
placed after it has been tested.
Solution
Since the events are dependent,
P(D1 and D2) = P(D,) * P(D2|D1)
= 2/25 * 1/24
= 2/600
= 1/300
Example 2:
The World Wide Insurance Company found that 53% of the residents of a city had homeowner's insurance with the company. Of these clients, 27% also had automobile insurance with the company.
If a resident is selected at random, find the probability that the resident has both homeowner's and automobile insurance with the World Wide insurance Company.
Solution:
P(H and A) = P(H) • P(A|H)
= (0.53) (0.27)
= 0.1431
This multiplication rule can be extended to three or more events, as shown in the
next example.
Example :
Three cards are drawn from an ordinary deck and not replaced. Find the probability of these.
a. Getting three jacks
b. Getting an ace, a king, and a queen in order
c. Getting a club, a spade, and a heart in order
d. Getting three clubs
Solution:
a. P(3 jacks) = 4/52 • 3/51 • 2/50
= 24/132,600
= 1/5525
b. P(ace and king and queen) = 4/52 • 4/51 • 4/50
= 64/132,600
= 8/16,575
c. P(club and spade and heart) = 13/52 •13/51 • 13/50
= 2197/132,600
= 169/10,200
d. P(3 clubs) = 13/52 • 12/51 • 11/50
= 1716/132,600
= 11/850
Tree diagrams can be used as an aid to finding the solution to probability problems when the events are sequential. Below Example illustrates the use of tree diagrams.
Example:
Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls an e red
ball. A coin is tossed. If it falls heads
up, box 1 is selected and a ball is drawn.If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selectng a red ball.
Solution:
With the use of a tree diagram, the sample space can be determined as shown in figure.First assign probabilities to each branch.Next, using the multiplication rule, multiply the probabilities for each branch.
Tree Diagram for the above example
Finally, use the addition rule, since a red ball can be obtained from box 1 or box 2.
P(Red) = 2/6 + 1/8
= 8/24 + 3/*24
= 11/24
(Note: The sum of all final probabilities will always be equal to 1.)
Tree diagrams can be used when the events are independent or dependent, and they
can also be used for sequences of three or more events.
Conditional probability:
Objective: Find the conditional probability of an event.
The conditional probability of an event B in relationship to an event A was defined as the probability that event B occurs after event A has already occurred.
The conditional probability of an event can be found by dividing both sides of the
equation for multiplication rule 2 by P(A), as shown:
P(A and B) = P(A) • P(B|A)
P(A and B)/P(A) = P(A) • P(B|A)/P(A)
P(A and B)/P(A) = P(B|A)
Formula for Conditional Probability
The probability that the second event B occurs given that the first event A has occurred Can be found by dividing the probability that both events occurred by the probability that the first event has occurred. The formula is
P(B|A) = P(A and B)/P(A)
Below examples illustrate the use of this rule.
Example 1 :
A box contains black chips and white chips. A person selects two chips without replacement. If the probability of selecting a black chip and a white chip is 15/56, and the probability of selecting a black chip on the first draw is 3/8 , find the probabilty of selecting white chip on the second draw, given that the first chip selected was a black chip.
Solution
Let
B = selecting a black chip W= selecting a white chip
Then
P(W|B) = P(B and W)/P(B)
= 15/56 ➗ 3/8
= 15/56 • 8/3
= 5/7
Hence, the probability of selecting a white chip on the second draw given that the first
chip selected was black is 5/7
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Example 2:
The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06,
and the probability that Sam cannot find a legal parking space and has to park in the no-
parking no-parking zone is 0.20. On Tuesday, Sam arrives at school and has to park in a
zone. Find the probability that he will get a parking ticket.
Solution:
Let
N= parking in a no-parking zone T= getting a ticket
Then
P(T|N) = P(N and T)/P(N)
= 0.06 / 0.20
= 0.30
Hence, Sam has a 0.30 probability of getting a parking ticket, given that he parked in a
no-parking Zone.
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The conditional probability of events occurring can also be computed when the data
are given in table form, as shown in below example.
Example:
A recent survey asked 100 people if they thought women in the armed forces should be
permitted to participate in combat. The results of the survey are shown in the table.
Gender Yes No Total
Male 32 18 50
Female 8 42 50
Total 40 60 100
Find these probabilities.
a. The respondent answered "yes, given that the respondent was a female.
b. The respondent was a male, given that the respondent answered "no."
Solution:
Let
M= respondent was a male Y=respondent answered "yes"
F= respondent was a female N=respondent answered "no"
a. The problem is to find P(Y|F). The rule states
P(Y|F) = P(F and Y)/P(F)
by the total number of respondents:
P(F and Y) = 8/100
The probability P(F) is the probability of selecting a female:
P(F) = 50/100
Then
P(Y|F) = P(F and Y)/P(F)
= 8/100 ➗ 50/100
= 8/100 • 100/50
= 4/25
b. The problem is to find P(M|N).
P(M|N) = P(N and M)/P(N)
= 18/100 ➗ 60/100
= 18/100 • 100/60
= 3/10
Venn diagram for conditional probability is shown in Figure 5-7. n this case,
P(B|A) = P(A and B)/P(A)
which is represented by the area in the intersection or overlapping part of the circles A and B divided by the area of Circle A. The reasoning here is that if one assumes A has occurred, then A becomes the sample space for the next calculation and is the denominator of probability fraction
P(A and B)/P(A).
The numerator P(A and B) represents the probability of the part of B that is contained in A. Hence, P(A and B) becomes the numerator of the probability fraction
P(A and B)/P(A).
Imposing a condition reduces the sample space
P(S)
P(B|A) = P(A and B)/P(A)
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