Here are a few recommended readings before getting started with this lesson.
Emily is a student at North High and top of her class. Her mathematics teacher asks her to select two students and form a team of three students for the coming year's mathematics competition. When Emily sees the student list, she notices that the four students who come after her have almost identical grades in math.
She decides to select two students out the four possibilities, which includes one boy and three girls. Use a sample space to determine whether the following events are dependent or independent.
To comprehend the probabilities of events in different situations, the following exploration can be used. Suppose there are three marbles — one blue, one green, and one orange — in a bowl.
In the following two cases, try to determine whether events A and B depend on each other or not.
Here the formal definitions of independent and dependent events will be presented.
Two events $A$ and $B$ are independent events if the occurrence of either of these events does not affect the occurrence of the other. It is also said that they are independent if and only if the probability that both events occur is equal to the product of the individual probabilities.
For example, consider drawing two marbles from a bowl, one at a time.
The probability of first picking a green marble can be calculated by dividing the favorable outcomes by the possible outcomes. Let $G,$ $B,$ and $O$ be the events of drawing green, blue, and orange marbles, respectively. There is $1$ green marble and $3$ marbles in total. $P(G)=31 $ Suppose that the first marble is replaced before the second draw. Therefore, after the replacement there is $1$ orange marble, and $3$ marbles in total. $P(O)=31 $ Note that there are $9$ possible outcomes for drawing two marbles one at a time. Only $1$ of these options corresponds to an event of drawing a green marble and then an orange marble. $GGGBGOBBBGBOOOOGOB $ Therefore, the combined probability of picking a green marble first and an orange marble second is $91 .$ Since the probability that both events occur is equal to the product of the individual probabilities, these events can be considered as independent events.
$31 ⋅31 =91 ⇓P(G)⋅P(O)=P(G∩O) $Two events $A$ and $B$ are considered dependent events if the occurrence of either of these events affects the occurrence of the other. If the events are dependent, the probability that both events occur is equal to the product of the probability of the first event occurring and the probability of the second event occurring after the first event.
For example, consider drawing two marbles from a bowl, one at a time.
This affects the probability of picking an orange marble on the second draw. Now there is still $1$ orange marble, but instead of $3,$ there are $2$ marbles in total. $P(O∣G)=21 $ Using this information, the sample space of the described situation can be found.
$GBGOBGBOOGOB $
Out of $6,$ there is only $1$ outcome that corresponds to first drawing a green marble and then an orange marble. Therefore, the probability of picking a green and then an orange marble is $61 .$ $31 ⋅21 =61 ⇓P(G)⋅P(O∣G)=P(G∩O) $
Because the occurrence of the first event affects the occurrence of the second, these events can be concluded to be dependent.Davontay wants to practice the independence of events. To do so, he uses a balanced six-sided die.
He rolls the die and considers the following events.
$ A={1,2}B={2,3,4,5,6} $
Based on the definitions of independent and dependent events, he tries to find out whether these events are independent or not. Help Davontay during his practice!
Start by calculating $P(A),$ $P(B),$ and $P(A∩B).$ Use the definition of independent events.
Recall that two events are independent if and only if the probability that both events occur is equal to the product of the individual probabilities. $P(A∩B)=P(A)⋅P(B) $
To calculate these probabilities, begin by identifying the sample space. Since Davontay rolls a six-sided die, there are $6$ elements in the sample space.
$S={1,2,3,4,5,6} $
Next, to find the probability that both events occur, identify the common elements in each event.
$ A={1,2 }B={2 ,3,4,5,6} $
Notice that there is only one element in common. Therefore, the number of favorable outcomes is $1$ and the number of possible outcomes is $6$ for $P(A∩B).$
$P(A∩B)=61 $
Now, the individual probabilities can be calculated proceeding in the same way. Since there are $2$ and $5$ elements in $A$ and $B,$ respectively, these are the number of favorable outcomes for each of these events. $ P(A)=62 P(B)=65 $
Finally, check whether these probabilities satisfy the definition of independent events.
Substitute values
Multiply fractions
$ba =b/2a/2 $
Since the probabilities do not satisfy the definition, event $A$ and event $B$ are not independent. This implies that $A$ and $B$ are dependent events.
Davontay decides that one exercise is not enough practice. He needs more practice with independent events. Given the probabilities that both events $A$ and $B$ occur, event $A$ occurs, and event $B$ occurs, help Davontay to decide whether $A$ and $B$ are independent or dependent events.
After Davontay is done practicing, he decides to watch the predictions of the semi-finals of Euro $2020.$ His favorite teams are England and Italy. He really wants to see these teams in the finals. According to the sports announcer, England has a $56%$ chance of winning against Denmark.
Davontay gets excited to hear the predictions for his other favorite team. However, the broadcast freezes for a couple of seconds and he misses the projection for Italy. He only hears that the chance of England playing against Italy is $20.72%.$
Davontay is quite curious about Italy's chances of winning against Spain. Keeping in mind that England's win does not affect Italy's win, help Davontay satisfy his curiosity.
Use the definition of independent events and the fact that England's and Italy's wins do not affect each other.
Let $E$ be the event of England winning against Denmark. Let $I$ be the event of Italy winning against Spain. From here, $P(E∩I)$ and $P(E)$ need to be identified. Note that $P(E∩I)$ is the probability that both England and Italy win against their opponents. Otherwise, they will not play against each other. For simplicity, convert the percentages to decimals. $P(E∩I)P(E) =0.2072=0.56 $ Since England's and Italy's wins do not affect each other, $E$ and $I$ are independent events. Therefore, the probability that both England and Italy win against their opponents is equal to the product of the probability that England wins and the probability that Italy wins.
$P(E∩I)=P(E)⋅P(I) $
Since $P(E∩I)$ and $P(E)$ are known, by substituting these values into this equation the probability that Italy wins against Spain can be found.$P(E∩I)=0.2072$, $P(E)=0.56$
$LHS/0.56=RHS/0.56$
Rearrange equation
Use a calculator
Davontay is excited to see his favorite teams in the Euro $2020$ finals. After the match, he goes to the library to study probability with his friend. Davontay wants to help his friend understand the independent and dependent events. To do this, he puts $8$ cards on a desk.
Davontay shuffles these cards and picks two cards, one at a time. Note that he replaces the first card before picking the second card.
Based on this, he defines two different cases for his friend. Help his friend answer these questions.
$P(A∩S)=P(A)⋅P(S) $ Davontay's friend should begin by calculating $P(A∩S),$ $P(A),$ and $P(S).$ Notice that there are $8$ cards and only one of them is the ace of spades. With this information, $P(A∩S)$ can be calculated by dividing the favorable outcomes by the possible outcomes.
$P(A∩S)=81 $
Using the same method, $P(A)$ and $P(S)$ can be also calculated.
$P(A)=82 P(S)=84 $
From here, by substituting these values into the equation it can be determined whether event $A$ and event $S$ are independent.
Substitute values
Multiply fractions
$ba =b/8a/8 $
Favorable Outcomes | Total Outcomes | Substitute | |
---|---|---|---|
$P(Q∩H)$ | $1$ | $8$ | $81 $ |
$P(Q)$ | $2$ | $8$ | $82 $ |
$P(H)$ | $2$ | $8$ | $82 $ |
Now that the probabilities have been calculated, Davontay's friend can find out whether the events are independent or dependent.
Substitute values
Multiply fractions
$ba =b/4a/4 $
After playing with the cards, Davontay's friend also wonders if the definition of independent events can be used to find one of the individual probabilities given that the events are independent. Therefore, Davontay provides several exercises for his friend, knowing that the definition of independent events is biconditional.
Let $A$ and $B$ be independent events. Given the probability that both event $A$ and event $B$ occur and the probability that event $A$ occurs, find the probability that event $B$ occurs.
This lesson has covered how to determine whether two events are independent or dependent. Using this knowledge, the challenge provided at the beginning of the lesson can be solved. Recall that Emily is selecting two students out of one boy and three girls who have the greatest math grades after Emily.
Use a sample space to determine whether the following events are dependent or independent.
Number of Girls | Outcome | |
---|---|---|
$111222 $ | $G_{1}BG_{2}BG_{3}BG_{1}G_{2}G_{1}G_{3}G_{2}G_{3} $ | $BG_{1}BG_{2}BG_{3}G_{2}G_{1}G_{3}G_{1}G_{3}G_{2} $ |
Therefore, there are $12$ outcomes in total. The events can be also identified as follows.
By the definition of independent events, if $P(A∩B)$ is the product of $P(A)$ and $P(B),$ then $A$ and $B$ are said to be independent events.
$P(A∩B)=P(A)⋅P(B) $ Analyzing the sample space, it can be seen that there are $9$ outcomes in which a girl is chosen first, $9$ outcomes in which a girl is chosen second, and $6$ outcomes in which two girls are chosen. Using this information, the values of $P(A),$ $P(B),$ and $P(A∩B)$ can be found. $P(A)P(B)P(A∩B) =129 =129 =126 $
Having found the probabilities, substitute these values into the equation.
Substitute values
$ba =b/6a/6 $
$ba =b/3a/3 $
Multiply fractions
Substitute values
$ba =b/3a/3 $
Multiply fractions