Translate the following events into set notation using the symbols A and B, complement, union, intersection. Also give the probability of the event as determined from the table above. Fill in the table below with these values. {If you cannot construct the symbol ∩, just use & or copy and paste} Lastly, draw a Venn diagram (on scratch paper, but you need not submit it) showing the events of Female, Public, and both Female and Public. The first one is completed for you.
Event in words Event in set notation Probability
Female and Public A ∩ B P(A ∩ B) = 0.38
Female and Private
Public and Male
Neither Female nor Public
c. Determine the probability that the student is Male. Do the same for student attending a Private high school. Use proper event notation remember that A is Female and B is Public.
A ∩ B
.
d. If you had not been given the table, but instead had merely been told that P(A)=.45 and P(B)=.84, would you have been able to calculate P(Ac) and P(Bc)? Explain how.
e. Calculate the probability that at least either the student is Female or the student attended a Public high school.
f. Another way to calculate P(A ∪ B) in part e is to use the complement of the event. First, in words what is the complement to the event described in e and then find this probability using the complement. Does this match your answer in e?
g. If you had not been given the table but instead had merely been told P(A)=.45 and P(B)=.84 and were asked to calculate P(A ∪ B), you might first consider P(A)+P(B). Calculate this sum and compare the result to your answer for e. Are they the same? Is this even a legitimate answer?
h. Given the knowledge that the student is Female, what is the conditional probability that the student attended a Public high school? [Hint: Restrict your consideration to Females, and ask yourself what fraction of those female students attended a Public high school?]
i. How does this conditional probability of the student attending a Public high school given that the student is Female compare with the unconditional probability of the student attending a Public high school? Does the knowledge that the student is Female make it more or less likely (or neither) that the student attended a Public high school?
this is the chart for this problem
Public Private Total
Female 38 7 45
Male 46 9 55
Total 84 16 100
P(B|A) = (P(A and B))/(P(A)) = (.38)/(.45) = 0.84
Event in words Event in set notation Probability
Female and Public A ∩ B P(A ∩ B) = 0.38
Female and Private A ∩ Bc
Public and Male B ∩ Ac
Neither Female nor Public Ac ∩ Bc
c. Determine the probability that the student is Male. Do the same for student attending a Private high school. Use proper event notation remember that A is Female and B is Public.
P(B) = P(Ac ∩ B) + P(A ∩ B) + P(A ∩ Bc)
So,
P(Ac ∩ B) = P(B) - P(A ∩ B) = 0.84 - 0.38 = 0.46
To calculate the probability of a student attending a private high school, we need to find P(A ∩ Bc).
P(A ∩ Bc) = P(A) - P(A ∩ B) = 0.45 - 0.38 = 0.07
d. If you had not been given the table, but instead had merely been told that P(A)=.45 and P(B)=.84, would you have been able to calculate P(Ac) and P(Bc)? Explain how.
Yes, we can calculate the complement probabilities.
P(Ac) = 1 - P(A) = 1 - 0.45 = 0.55
P(Bc) = 1 - P(B) = 1 - 0.84 = 0.16
e. Calculate the probability that at least either the student is Female or the student attended a Public high school.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 0.45 + 0.84 - 0.38 = 0.91
f. Another way to calculate P(A ∪ B) in part e is to use the complement of the event. First, in words, what is the complement to the event described in e, and then find this probability using the complement. Does this match your answer in e?
The complement to the event "at least either the student is Female or the student attended a Public high school" is "Neither Female nor Public."
P(Ac ∩ Bc) = 1 - P(A ∪ B) = 1 - 0.91 = 0.09
The answer in part e matches with this calculated probability.
g. If you had not been given the table but instead had merely been told P(A)=.45 and P(B)=.84 and were asked to calculate P(A ∪ B), you might first consider P(A)+P(B). Calculate this sum and compare the result to your answer for e. Are they the same? Is this even a legitimate answer?
P(A) + P(B) = 0.45 + 0.84 = 1.29
Since the sum of probabilities cannot exceed 1, this is not a legitimate answer. The correct probability of P(A ∪ B) is 0.91, as calculated in part e.
h. Given the knowledge that the student is Female, what is the conditional probability that the student attended a Public high school? [Hint: Restrict your consideration to Females, and ask yourself what fraction of those female students attended a Public high school?]
The conditional probability that a student attended a Public high school (B) given that the student is Female (A) can be calculated using the formula:
P(B|A) = P(A ∩ B) / P(A)
P(B|A) = 0.38 / 0.45 ≈ 0.8444
i. How does this conditional probability of the student attending a Public high school given that the student is Female compare with the unconditional probability of the student attending a Public high school? Does the knowledge that the student is Female make it more or less likely (or neither) that the student attended a Public high school?
The conditional probability P(B|A) ≈ 0.8444 is very close to the unconditional probability of a student attending a Public high school P(B) = 0.84.
Knowledge that the student is Female does not significantly change the likelihood that the student attended a Public high school since the conditional probability is approximately the same as the unconditional probability.
To solve these questions, we will use the set notation and the given symbols A and B, complement, union, and intersection.
a. Event in set notation:
1. Female and Public: A ∩ B
The probability of this event is given as P(A ∩ B) = 0.38.
2. Female and Private: A ∩ B'
We use the complement operator (') to represent Private, so A ∩ B' represents the event of being Female and attending a Private high school.
3. Public and Male: A' ∩ B
Similarly, A' ∩ B represents the event of being Male and attending a Public high school.
4. Neither Female nor Public: A' ∩ B'
A' ∩ B' represents the event of being neither Female nor Public.
b. Probability of being Male and attending a Private high school:
P(A' ∩ B') will give us the probability.
c. If we only had P(A) = 0.45 and P(B) = 0.84, we could calculate the complements P(Ac) and P(Bc) using the formula:
P(Ac) = 1 - P(A) = 1 - 0.45 = 0.55
P(Bc) = 1 - P(B) = 1 - 0.84 = 0.16
d. To calculate the probability that at least either the student is Female or the student attended a Public high school (A ∪ B):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
e. Another way to calculate P(A ∪ B) is to use the complement. The complement to the event described in e is the event of being neither Female nor Public (A' ∩ B').
P(A' ∩ B') = 1 - P(A ∪ B)
f. To compare the results in e and f, we can calculate P(A ∪ B) using P(A) + P(B) - P(A ∩ B) and compare it with 1 - P(A ∪ B).
g. If we had P(A) = 0.45 and P(B) = 0.84, we can calculate P(A ∪ B) using P(A) + P(B). The result might not match the answer in e because it doesn't consider the intersection of A and B (P(A ∩ B)).
h. Given the knowledge that the student is Female, the conditional probability of the student attending a Public high school can be found using the formula:
P(B | A) = P(A ∩ B) / P(A)
i. To compare the conditional probability of the student attending a Public high school given that the student is Female with the unconditional probability of the student attending a Public high school, we can compare P(B | A) with P(B). This will determine if the knowledge of being Female affects the likelihood of attending a Public high school.