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| 8 Theory slides |
| 6 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
The remaining results from the survey are organized in the following table.
Consider the presented data to find the probabilities of the following scenarios.
Substitute values
c/da/b=ba⋅cd
Multiply fractions
ba=b/5a/5
Substitute values
c/da/b=ba⋅cd
Multiply fractions
ba=b/15a/15
The probability that the second book is a Geometry book given that the first book chosen is a History book equals P(H)P(H∩G). |
The previous statement can also be rewritten in terms of H and G as follows.
The probability that event G happens given that event H happened equals P(H)P(H∩G). |
Similarly, the second probability found in part B equals the probability found in part D. This leads to write the following relation.
The probability that the first book is a History book, given that the second book is a Geometry book equals P(G)P(H∩G). |
As before, the previous statement can be rewritten in terms of H and G.
The probability that event H happens, given that event G happened equals P(G)P(H∩G). |
Considering these details, it can be concluded that 2% of the bags containing forbidden items could trigger the alarm and 96% of the bags that do not have forbidden items could not trigger the alarm.
Now, using the percentages in the branches, the number of bags for each event can be found.
Forbidden and Alarm | 350⋅98%=343 |
---|---|
Forbidden and No Alarm | 350⋅2%=7 |
Not Forbidden and Alarm | 4650⋅4%=186 |
Not Forbidden and No Alarm | 4650⋅96%=4464 |
Finally, all the information can be shown on the tree diagram.
Calculate quotient
Convert to percent
Probabilities of the Events | ||
---|---|---|
P(Alarm and Forbidden) | 5000343=6.86% | |
P(No Alarm and Forbidden) | 50007=0.14% |
Take note that the sum of the probabilities is equal to 7%, which is the percentage of the bags that contain forbidden items.
Alarm.
Calculate quotient
Round to 4 decimal place(s)
Convert to percent
Calculate quotient
Round to 4 decimal place(s)
Convert to percent
There is about a 64.84% chance that Mark's bag contains a forbidden item. |
This probability is not close enough to 100% to ensure that Mark's bag contains a forbidden item. Therefore, it is doubtful — but possible — that Mark's bag contains a forbidden item. Next, recall the answer found in Part D.
There is about a 0.16% chance that Izabella's bag contains a forbidden item. |
Since this probability is very small — less than 1% — it is almost certain that Izabella does not have forbidden items in her bag — but still possible.
Conditional probability is the measure of the likelihood of an event B occurring, given that event A has occurred previously. The probability of B given A is written as P(B∣A). It can be calculated by dividing the probability of the intersection of A and B by the probability of A.
P(B∣A)=P(A)P(A and B),where P(A)=0
The intuition behind the formula can be visualized by using Venn Diagrams. Consider a sample space S and the events A and B such that P(A)=0.
Assuming that event A has occurred, the sample space is reduced to A.
This means that the probability that event B can happen is reduced to the outcomes in the intersection of A and B, that is, to those outcomes in A∩B.
The possible outcomes are given by P(A) and the favorable outcomes by P(A∩B). Therefore, the conditional probability formula can be obtained using the probability formula.
P(B∣A)=P(A)P(A and B)
Diego's generous father has finished doing laundry and put Diego's T-shirts along with those of his big brother into the same ol’ basket. There are orange, blue, and red T-shirts in the basket, of which four are S-sized and eight are M-sized.
P(S and O)=121, P(O)=125
c/da/b=ba⋅cd
Multiply fractions
ba=b/12a/12
P(S and O)=121, P(S)=31
c/da/b=ba⋅cd
Multiply fractions
ba=b/3a/3
P(S and B)=61, P(B)=31
c/da/b=ba⋅cd
Multiply fractions
ba=b/3a/3
P(S and B)=61, P(S)=31
c/da/b=ba⋅cd
Multiply fractions
ba=b/3a/3
Find the required conditional probability and round it to two decimal places.
To find the corresponding probabilities, take a look at the table.
P(Penguin and France)=381, P(France)=190157
c/da/b=ba⋅cd
Multiply fractions
Calculate quotient
Convert to percent
Round to nearest integer
P(Penguin and France)=381, P(Penguins)=387
c/da/b=ba⋅cd
Multiply fractions
Calculate quotient
Convert to percent
Round to nearest integer
P(Antarctica and No Penguin)=1903, P(No Penguin)=3831
c/da/b=ba⋅cd
Multiply fractions
Calculate quotient
Convert to percent
Round to nearest integer
A jar contains blue and red marbles.
To calculate P(large|red), we will use the following formula. P( A| B)=P( A and B)/P( B) If we apply this formula to our situation, we get the following equation. P(large|red)=P( large and red)/P(red) The probability of drawing a red marble is the ratio of the number of red marbles to the total number of marbles. Examining the jar, we count 6 red marbles and 10 marbles in total. Now we can determine P(red).
Let's also calculate the probability of drawing a large marble that is also red. We have 2 large red marbles. With this information, we can calculate P(large and red).
Now we have everything we need to calculate P(large|red).
As in the previous part, we must determine two probabilities in order to calculate the given conditional probability. The jar contains 4 blue marbles. We also see that 2 of the blue marbles are large.
P(blue and large)&=2/10=1/5 [0.7em]
P(large)&=4/10=2/5
Now we can calculate the given conditional probability.
Here we have two conditional probabilities we must determine.
P(red|small)&=P(red and small)/P(small) [2em]
P(small|red)&=P(small and red)/P(red)
Note that the numerators of both fractions on the right-hand side of the equations are equivalent. We already know the probability of a marble being red is 35. We have 6 small marbles, of which 4 are red. With this information, we can determine the remaining probabilities.
P(red and small)&=4/10=2/5 [1em]
P(small)&=6/10=3/5
Now we have all the information we need to determine both probabilities.
Let's next determine the second probability.
As we can see, the probabilities are the same. Note that this did not have to be the case though. There is nothing that states that P(B|A) must be equal to P(A|B).
Jordan is employed as a mechanic. Every month she does yearly inspections of cars. The following table shows the result of a month's work.
Use the information to calculate the following conditional probabilities. Answer with a fraction in its simplest form.
The expression P(pass|defective) is the probability of a car passing the inspection given that it was defective. To determine this, we must use the following formula. P( A| B)=P( A and B)/P( B) If we apply this formula to our given situation, we get the following equation.
To determine the probabilities in the numerator and denominator of the fraction, we have to divide the number of cars that fits the given descriptions by the total number of cars. To do this, let's first calculate the total number of cars inspected.
By adding the all the numbers in the table we can determine the total number of cars that were inspected during the month. 2+ 40+ 450+ 10=502 Now we can find each probability.
Let's next determine the probability of a car getting a pass and being defective.
Now we can determine the conditional probability.
To calculate this conditional probability, we must divide the probability of a car failing the inspection and being non-defective by the probability that a car is non-defective.
450+ 10=460
From the table, we see that 460 cars were non-defective.
Let's also calculate the probability of a car both failing the inspection and being non-defective. From the table, we see that 10 cars failed the inspection while being non-defective.
Now we can determine the given conditional probability.
The expression P(x<10|x<13) is the probability of an observation in the sample space being less than 10, given that the observation is less than 13. To calculate this, we will use the following formula. P( A| B)=P( A and B)/P( B) If we apply this formula to the given situation, we get the following equation. P( x < 10| x < 13) = P( x < 10 and x < 13)/P( x < 13) We need to determine the probabilities in the numerator and denominator of the fraction. We have a total of ten observations and eight of them are below 13.
Let's also calculate the second probability.
Now we can determine the given conditional probability.
This time, we need to determine the probability of an observation being less than thirteen as well as the probability of an observation being greater than or equal to seven and less than thirteen. From the previous section we know the probability of a number being less than thirteen.
P(x<13)=4/5
Examining the diagram, we see that six observations are greater than or equal to seven and also less than thirteen. With this information, we can determine the probability.
Now we can determine the given conditional probability.