Respuesta :
Step-by-step explanation:
(a) What is the probability that the selected student has at least one of the three types of cards?
Here we have:
P(A) = 0.6 is the probability that the student has a Visa card
P(B) = 0.4 is the probability that the student has a MasterCard
P(C) = 0.2 is the probability that the student has an American Express card
Here we want to find
P(A U B U C)
which is the probability that the student has at least one of the 3 cards.
This can be calculated as:
P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C)
Where:
P(A ∩ B) = 0.3,
P(A ∩ C) = 0.12
P(B ∩ C) = 0.1
And substituting,
P(A U B U C) = 0.6 + 0.4 + 0.2 - 0.3 - 0.12 - 0.1 = 0.68
(b) What is the probability that the selected student has both a Visa card and a MasterCard but not an American Express card?
The probability that the student has both a Visa card (event A) and a MasterCard (event B) is given by
P(A ∩ B) = 0.3
Also we know that the probability that the student has an American Express card (event C) is
P(C) = 0.2
Therefore, the probability that the student does NOT have an American Express card is
P(C') = 1 - P(C) = 1 - 0.2 = 0.8
Therefore, the probability that the selected student has both a Visa card and a MasterCard but NOT an American Express card is the intersection between (A ∩ B) and the set (C'), which is therefore:
P(A ∩ B ∩ C') = P(A ∩ B) - P(A ∩ B ∩ C) = 0.3 - 0.07 = 0.23
(c) Calculate P(B | A) and P(A | B).
From the definition of conditional probability, we have:
[tex]P(B | A) = \frac{P(A\cap B)}{P(A)}[/tex]
where in this problem,
P(A ∩ B) = 0.3
P(A) = 0.6
Substituting,
[tex]P(B | A) = \frac{0.3}{0.6}=0.5[/tex]
Which is the probability that a student has a MasterCard (event B) given that he also has a Visa Card (event A).
Moreover,
[tex]P(A| B) = \frac{P(A\cap B)}{P(B)}[/tex]
and given
P(A ∩ B) = 0.3
P(B) = 0.4
We get
[tex]P(A|B) = \frac{0.3}{0.4}=0.75[/tex]
Which is the probability that a student has a Visa Card (event A) given that he also has a Master Card (event B).
(d) Interpret P(B | A) and P(A | B). (Select all that apply.)
As we said, the interpretation of the conditional probability is:
P(X | Y) is the probability that event X occurs given that event Y has already occurred.
As we said in part c, the correct answers of the conditional probabilities are:
P(B | A) is the probability that given that a student has a Visa card, they also have a MasterCard.
P(A | B) is the probability that given that a student has a Master card, they also have a VisaCard.
(e) If we learn that the selected student has an American Express card, what is the probability that she or he also has both a Visa card and a MasterCard?
In this problem, we are basically asked to find
P(A ∩ B | C)
which is the probability that, given that the student has an American Express card (event C), then he also has both a Visa Card and a MasterCard (event A ∩ B)
Applying the definition of conditional probability,
[tex]P(A \cap B | C) = \frac{P(A \cap B \cap C)}{P(C)}[/tex]
where we have:
P(A ∩ B ∩ C) = 0.07
P(C) = 0.2
Substituting,
[tex]P(A \cap B | C) = \frac{0.07}{0.2}=0.35[/tex]
(f) Given that the selected student has an American Express card, what is the probability that she or he has at least one of the other two types of cards?
In this problem, we are asked to find
P(A U B | C)
which is the probability that, given that the student has an American express card (event C), then he/she has at least one between Visa Card and Master Card (event A U B)
By applying the definition,
[tex]P(A \cup B | C) = \frac{P((A \cup B) \cap C)}{P(C)}[/tex]
Here we need to find [tex]P((A \cup B) \cap C)[/tex]. This is actually the intersection of (A or B) with C: therefore, it is equal to the sum of the intersections (A and C) and (B and C) minus the intersection of the three events (A ∩ B ∩ C), so
[tex]P((A \cup B) \cap C) = P(A\cap C) + P(B\cap C)-P(A\cap B \cap C)=0.12+0.1-0.07=0.15[/tex]
Therefore, given that
P(C) = 0.2
We have
[tex]P(A \cup B | C) = \frac{0.15}{0.20}=0.75[/tex]
