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In this lesson the *Multiplication Rule of Probability* will be introduced. Also, this lesson will explore how this rule can be used to find the probability of the intersection of two events and how it can be used to reverse conditional probability.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Consider the following two events.

Event A | The DNA from a crime scene is concluded to be the DNA of the defendant. |
---|---|

Event B | The defendant is not guilty. |

$P(A∣B)is small⇒?P(B∣A)is small $

This reasoning is based on the assumption that $P(A∣B)$ is equal to $P(B∣A).$ Is this assumption, and therefore the reasoning for it, correct or incorrect? Why?
The probability of being infected with a certain disease is 0.0001. There is a test used to detect this disease. If someone is infected, the test comes back positive 97 times out of 100 cases. If someone is not infected, the test comes back positive 1 time out of 10000 cases.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["0.5"]}}

$P(infected)P(positive∣infected)P(positive∣not infected) =0.0001=0.97=0.0001 $

Express the answers as decimal numbers approximated to one significant figure. a What is the probability that a person is infected when they have a negative test result?

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["0.000003"]}}

b What is the probability that a person is actually infected when they have a positive test result?

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["0.5"]}}

c Finally, what is the probability that a person is not infected when they have a positive test result?

Conversely, if A and B are dependent events, a rearrangement of the Conditional Probability Formula can be used to find the probability of the intersection of the events.

$P(AandB)=P(A)⋅P(B∣A)orP(AandB)=P(B)⋅P(A∣B) $

Consider two dependent events A and B. The conditional probability of A given B is the ratio of the probability of the intersection of A and B to the probability of B.
The Multiplication Rule of Probability is obtained by multiplying both sides of the above formula by P(B) and using the Symmetric Property of Equality.
Following similar reasoning, an equivalent form of the rule can be obtained.

Consider two sets of integer numbers.
A number a is chosen from the union of C and D. What is the probability that the number is a positive element of C? What is the probability that the number is a negative element of D? Use a tree diagram to determine the probabilities. ### Hint

### Solution

Now the probabilities will be added to the diagram.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mspace\" style=\"margin-right:0.16666666666666666em;\"><\/span><span class=\"minner\"><span class=\"mopen delimcenter\" style=\"top:0em;\">(<\/span><span class=\"mord mathdefault\">a<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2208<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">C<\/span><span class=\"mord text\"><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">and<\/span><span class=\"mord\">\u00a0<\/span><\/span><span class=\"mord mathdefault\">a<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mclose delimcenter\" style=\"top:0em;\">)<\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["\\dfrac{3}{10}","0.3"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mspace\" style=\"margin-right:0.16666666666666666em;\"><\/span><span class=\"minner\"><span class=\"mopen delimcenter\" style=\"top:0em;\">(<\/span><span class=\"mord mathdefault\">a<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2208<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">D<\/span><span class=\"mord text\"><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">and<\/span><span class=\"mord\">\u00a0<\/span><\/span><span class=\"mord mathdefault\">a<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\"><<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mclose delimcenter\" style=\"top:0em;\">)<\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["\\dfrac{1}{10}","0.1"]}}

Outcomes that should be included in the tree diagram are $a∈C,$ $a∈D,$ a>0, and a<0.

A tree diagram will be made to answer the questions. The root node of the tree represents the event of choosing a number a from $C∪D.$ The chosen number can be either from C or from D.

The probability of each outcome should be written on the corresponding branch. Set C has 6 elements, D has 4 elements, and the sets do not have any elements in common. With this information the probability that a is an element of C and that a is an element of D can be calculated.The other outcomes that need to be considered to answer the questions are that a can be positive or negative.

Set C has 6 elements. 3 of them are positive and 3 of them are negative. Therefore, knowing that a is an element of C, the probability that it is positive is $63 $ and the probability it is negative is also $63 .$

Likewise, D has 4 elements: 3 of them positive and 1 of them negative. Knowing that a is from D, the probability it is positive is $43 $ and the probability it is negative is $41 .$

The probability that a is an element of C *and* positive can be calculated by multiplying the probabilities along the corresponding branch of the tree diagram.

Similarly, the probability that a is an element of D *and* negative can be found.

For the following questions, approximate the answers to two decimal places.

Dylan participates in a school lottery that consists of two stages. He starts by drawing a ticket from a hat. This ticket tells Dylan from which of three boxes he will be drawing a second ticket. There are 30 tickets in the hat: 12 for box A, 10 for box B, and 8 for box C.

Each box has 5 tickets. Box A has 1 winning ticket, box B has 2 winning tickets, and box C has 3. What is the probability that Dylan wins? Express the answer as a percent rounded to the nearest whole number.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">Dylan<\/span><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">wins<\/span><\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mord\">%<\/span><\/span><\/span><\/span>","answer":{"text":["37"]}}

Use a tree diagram to represent the given information.

First make a tree diagram. The root node of the tree represents the event of drawing a ticket from the hat. There are 30 tickets in the hat. From those, $12$ correspond to box A, 10 correspond to box B, and 8 correspond to box C. With this information, the probabilities for the first three branches of the tree can be written.

A ticket drawn from any of the boxes can be a winning or a loosing ticket. Each box contains 5 tickets. Box A has $1$ winning ticket, box B has 2, and box C 3. With this information, the probabilities of drawing a winning ticket from each box can be written.

The sum of the probabilities of the branches that come out of the same node is equal to 1. Knowing this, the probabilities of not drawing a winning ticket from each box can be calculated.

There are three possible outcomes in which Dylan draws a winning ticket.

The sum of the probabilities of these outcomes is the probability that Dylan draws a winning ticket.P(win)=P(box A)⋅P(win∣box A)+P(box B)⋅P(win∣box B)+P(box C)⋅P(win∣box C)

SubstituteValues

Substitute values

$P(win)=3012 ⋅51 +3010 ⋅52 +308 ⋅53 $

Evaluate right-hand side

ReduceFrac

$ba =b/6a/6 $

$P(win)=52 ⋅51 +3010 ⋅52 +308 ⋅53 $

ReduceFrac

$ba =b/10a/10 $

$P(win)=52 ⋅51 +31 ⋅52 +308 ⋅53 $

ReduceFrac

$ba =b/2a/2 $

$P(win)=52 ⋅51 +31 ⋅52 +154 ⋅53 $

MultFrac

Multiply fractions

$P(win)=252 +152 +7512 $

ExpandFrac

$ba =b⋅3a⋅3 $

$P(win)=756 +152 +7512 $

ExpandFrac

$ba =b⋅5a⋅5 $

$P(win)=756 +7510 +7512 $

AddFrac

Add fractions

$P(win)=7528 $

FracToDiv

$ba =a÷b$

$P(win)=0.373333…$

WritePercent

Convert to percent

$P(win)=37.3%$

RoundInt

Round to nearest integer

$P(win)≈37%$

Consider two boxes A and B that both contain orange and purple marbles. A box is randomly chosen and then one marble is drawn from this box. The following probabilities are known.
### Hint

### Solution

$P(orange)P(box A)P(orange∣box A) =125 =127 =73 $

Knowing that the drawn marble is orange, what is the probability that it was taken from box A? {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["3\/5","\\dfrac{3}{5}","0.6"]}}

How can the probability that the marble is from box A and it is orange be calculated using the given information?

The probability that a marble is from box A knowing that it is orange is a conditional probability. It is the *reverse* of the given conditional probability.
The probability that the marble is from box A if it is orange equals $53 .$

$DesiredP(box A∣orange) GivenP(orange∣box A) $

By the definition of a conditional probability, the desired conditional probability can be expressed as follows.
$P(box A∣orange)=P(orange)P(box A and orange) $

The probability in the denominator is known. However, the probability in the numerator is not. Since the expression in the numerator is the probability of the intersection of the events marble from box Aand

orange marble,it can be rewritten using the Multiplication Rule of Probability.

$P(box A and orange)=P(orange∣box A)⋅P(box A) $

The above expression can be substituted into the formula for the probability that the marble is from box A knowing it is orange.
$P(box A∣orange)=P(orange)P(box A and orange) ⇕P(box A∣orange)=P(orange)P(orange∣box A)⋅P(box A) $

The conditional probability has been expressed in terms of the known probabilities. Now, the values of the known probabilities can be substituted into the formula.
$P(box A∣orange)=P(orange)P(orange∣box A)⋅P(box A) $

SubstituteValues

Substitute values

$P(box A∣orange)=125 73 ⋅127 $

Evaluate right-hand side

MultFrac

Multiply fractions

$P(box A∣orange)=125 7⋅123⋅7 $

CancelCommonFac

Cancel out common factors

$P(box A∣orange)=125 7 ⋅123⋅7 $

SimpQuot

Simplify quotient

$P(box A∣orange)=5/123/12 $

DivFracByFracSameDenom

$b/12a/12 =ba $

$P(box A∣orange)=53 $

CalcQuot

Calculate quotient

P(box A∣orange)=0.6

For the following questions, approximate the answers to two decimal places.

The following probabilities are known.
### Solution

### Probability That a Person Is Infected if the Test Is Negative

By the definition of conditional probability, the conditional probability that a person is infected given their test result is negative can be written as a ratio.
### Probability That a Person Is Infected if the Test Is Positive

The probability that a person is infected if the test is positive can be written as a ratio.
Finally, the desired conditional probability will be found.
### Probability That a Person Is Not Infected if the Test Is Positive

The probability that a person is not infected given the test result is positive can be found by reversing the conditional probability.
### Extra

Comparing Reversed Conditional Probabilities

$P(infected)P(positive∣infected)P(positive∣not infected) =0.0001=0.97=0.0001 $

Using this information, calculate three other probabilities. Write the answers correct to one significant digit. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">P<\/span>(<span class=\"text\">infected<\/span>\u2223<span class=\"text\">negative<\/span>)<span class=\"space big-before big-after\">=<\/span><\/span>","formTextAfter":null,"answer":{"text":["0.000003"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">P<\/span>(<span class=\"text\">infected<\/span>\u2223<span class=\"text\">positive<\/span>)<span class=\"space big-before big-after\">=<\/span><\/span>","formTextAfter":null,"answer":{"text":["0.5"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">P<\/span>(<span class=\"text\">not infected<\/span>\u2223<span class=\"text\">positive<\/span>)<span class=\"space big-before big-after\">=<\/span><\/span>","formTextAfter":null,"answer":{"text":["0.5"]}}

First, the given information will be represented in a tree diagram.

The sum of the probabilities of branches coming out of the same node is always 1. Using this fact, the tree diagram can be completed.

The tree diagram gathers all of the known information. Now, the desired unknown probabilities will be calculated one by one.

$P(infected∣negative)=P(negative)P(infected∩negative) $

The probability of the intersection of two events can be rewritten using the Multiplication Rule of Probability.
$P(infected∣negative)=P(negative)P(negative∣infected)⋅P(infected) $

Both probabilities in the numerator are represented in the tree diagram.
The probability that the test comes back negative can be found using the tree diagram as well. To do so, the probabilities of all outcomes that include receiving a negative test result should be added. There are two such outcomes. These are infected with a negative test result

and not infected with a negative test result

.

$P(infected∣negative)=P(negative)P(negative∣infected)⋅P(infected) $

SubstituteValues

Substitute values

$P(infected∣negative)=0.999803010.03⋅0.0001 $

Evaluate right-hand side

Multiply

Multiply

$P(infected∣negative)=0.999803010.000003 $

UseCalc

Use a calculator

$P(infected∣negative)=0.0000030005…$

RoundSigDig

Round to 1 significant

$P(infected∣negative)≈0.000003$

$P(infected∣positive)=P(positive)P(infected∩positive) $

Again, the probability in the numerator can be rewritten using the Multiplication Rule of Probability.
$P(infected∣positive)=P(positive)P(positive∣infected)⋅P(infected) $

It is known that P(infected)=0.0001 and P(positive∣infected)=0.97. Now, the probability that the test result is positive has to be found. Since the result of the test can be either positive or negative, the probability that it is positive is the difference of 1 and the probability that the result is negative.
P(positive)=1−P(negative)

Substitute

$P(negative)=0.99980301$

P(positive)=1−0.99980301

SubTerm

Subtract term

$P(positive)=0.00019699$

$P(infected∣positive)=P(positive)P(positive∣infected)⋅P(infected) $

SubstituteValues

Substitute values

$P(infected∣positive)=0.000196990.97⋅0.0001 $

Evaluate right-hand side

Multiply

Multiply

$P(infected∣positive)=0.000196990.000097 $

UseCalc

Use a calculator

$P(infected∣positive)=0.49241…$

RoundSigDig

Round to 1 significant

$P(infected∣positive)≈0.5$

$P(not infected∣positive)=P(positive)P(positive∣not infected)⋅P(not infected) $

It is known that $P(positive)=0.00019699.$ The remaining two probabilities in the above expression are represented on the tree diagram.
All of the information needed to calculate the conditional probability has been found.
$P(not infected∣positive)=P(positive)P(positive∣not infected)⋅P(not infected) $

SubstituteValues

Substitute values

$P(not infected∣positive)=0.000196990.0001⋅0.9999 $

Evaluate right-hand side

Multiply

Multiply

$P(not infected∣positive)=0.000196990.00009999 $

UseCalc

Use a calculator

$P(not infected∣positive)=0.50758…$

RoundSigDig

Round to 1 significant

$P(not infected∣positive)≈0.5$

The two given conditional probabilities have been *reversed*. Based on that, the example values of P(A∣B) and P(B∣A) can be compared.

P(positive|infected) | P(infected|positive) |
---|---|

0.97 | 0.5 |

P(positive|not infected) | P(not infected|positive) |

0.0001 | 0.5 |

It is seen that P(A∣B) and P(B∣A) are not equal.

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