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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
On the face of it, one may think that it is equally likely that Joe has a brother or sister, suggesting that the probability of the other child being a boy is 50 %. However, this is not correct. From the exercise, we need to find the probability that both children are boys given that at least one of them is a boy. This is a conditional probability. P(Both boys|At least one boy) = ? Consider a random family with two children. We have four possible sibling constellations depending on which child was born first.
Firstborn | Secondborn | Event |
---|---|---|
Boy (B) | Boy (B) | { B,B } |
Boy (B) | Girl (G) | { B,G } |
Girl (G) | Boy (B) | { G,B } |
Girl (G) | Girl (G) | { G,G } |
Of the four events, three of them contain at least one boy. Additionally, there is only one event where the two children are boys. With this information, we can write down the probabilities we need. P(Both boys) &= 1/4 [0.7em] P(At least one Boy) &= 3/4 Now we can determine the conditional probability that Kevin's neighbor has two boys given that one of her kids is a boy.
The probability that both children are boys is 13.