| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount }} |
| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount }} |
| {{ 'ml-lesson-time-estimation' | message }} |
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. |
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.
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.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.Substitute values
ba=b/6a/6
ba=b/10a/10
ba=b/2a/2
Multiply fractions
ba=b⋅3a⋅3
ba=b⋅5a⋅5
Add fractions
ba=a÷b
Convert to percent
Round to nearest integer
How can the probability that the marble is from box A and it is orange be calculated using the given information?
marble from box Aand
orange marble,it can be rewritten using the Multiplication Rule of Probability.
Substitute values
Multiply fractions
Cancel out common factors
Simplify quotient
b/12a/12=ba
Calculate quotient
For the following questions, approximate the answers to two decimal places.
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.
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
.
Substitute values
Multiply
Use a calculator
Round to 1 significant digit(s)
P(negative)=0.99980301
Subtract term
Substitute values
Multiply
Use a calculator
Round to 1 significant digit(s)
Substitute values
Multiply
Use a calculator
Round to 1 significant digit(s)
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.