Another probability question!

[scresthop123]

Consider randomly selecting a student at a ceratin university, and let A denote the event that the selected individual has a Visa credit card and B be the analogous event for a MAsterCard. Suppose that P(A)= .5 and P(B)= .4 and P(A intersection B) = .25.

A. Compute the probability that the selected individual has at least one of the two types of cards.
B. What is the probability that the selected individual has neither type of card?
C. Describe, in terms of A and B, the event that the selected student has a Visa Card but not a MAsterCard, and then calculate the probability of this event.
D. Calculate P(B/A), P(A/B), P(complement of A/B), and P(complement of B/A) and describe what each represents.
E. Given that the selected individual has at least one card, what is the probability that he or she has a Visa Card?

Help with all or any, please!!!!

Thanks!

 

[tkhunny]

Draw to intersecting circles.
Label one "Visa" and the other "MasterCard"
Put 0.25 in the intersection.
In the non-intersection portion of the "Visa" circle, put 0.50 - 0.25 = 0.25
In the non-intersection portion of the "MasterCard" circle, put 0.40 - 0.25 = 0.15
In the outer area, in neither circle, put 1.00 - 0.50 - 0.40 + 0.25 = 0.35

Check 0.25 + 0.25 + 0.15 + 0.35 = 1.00 Check!

Now answer all the questions.

 

[soroban]

Hello, scresthop123!

Randomly select a student at a certain university.

Let \(\displaystyle V\) = the student has a Visa card, and \(\displaystyle M\) = the student has a MasterCard.

Suppose that: .\(\displaystyle P(V)\,=\, 0.5,\;\;P(M)\,=\, 0.4,\;P(V \cap M) \,=\,0.25\)


A. Find the probability that the student individual has at least one of the two types of cards.

\(\displaystyle \text{Formula: }\(V \cup M) \;=\;P(V) + P(M) - P(V \cap M)\)

\(\displaystyle \text{Therefore: }\;P(V \cup M) \;=\;0.5 + 0.4 - 0.25 \;=\;0.65\)

B. What is the probability that the selected individual has neither type of card?

This is the exact opposite of part A.

\(\displaystyle P(V' \cap M') \;=\;1 - 0.65 \;=\;0.35\)

C. Describe, in terms of V and M, the event that the selected student has a Visa Card but not a MasterCard,
. . . and then calculate the probability of this event.

\(\displaystyle P(V \cap M') \;=\;P(V) - P(V \cap M) \;=\;0.5 - 0.25 \;=\;0.25\)

D. Calculate \(\displaystyle P(M|V),\; P(V|M),\; P(V'|M),\; P(M'|V)\), and describe what each represents.

\(\displaystyle \text{We will use Bayes' Theorem: }\;P(A|B) \;=\;\frac{PA \cap B)}{P(B)}\)


\(\displaystyle P(M|V) \;=\;\frac{P(M \cap V)}{P(V)} \;=\;\frac{0.25}{0.5} \;=\;0.5\)

This is the probability that the student has a MasterCard, given that he has a Visa.


\(\displaystyle P(V|M) \;=\;\frac{P(V \cap M)}{P(M)} \;=\;\frac{0.25}{0.4} \;=\;0.625\)

This is the probability that the student has a Visa, given that he has a MasterCard.


\(\displaystyle P(V'|M) \;=\;\frac{P(V' \cap M)}{P(M)} \;=\;\frac{0.15}{0.4} \;=\;0.375\) .**

This is the probability that the student does not have a Visa, given that he has a MasterCard.


\(\displaystyle P(M'|V) \;=\;\frac{P(M' \cap V)}{P(V)} \;=\; \frac{0.25}{0.5} \;=\;0.5\) .**

This is the probability that the studehnt does not have a MasterCard, given that he has a Visa.

E. Given that the student has at least one card, what is the probability that he/she has a Visa Card?

\(\displaystyle \text{We want: }\;P(V\,|\text{ at least one card}) \;=\;\frac{P(V \cap \text{at least one card})}{P(V \cup M)}\)


\(\displaystyle \text{Numerator: }\(V \cap [V \cup M])\)
. . This is the probability that he has a Visa and has at least one card.

. . \(\displaystyle P(V) \:=\:0.5\)

. . But if he has Visa and then he has at least one card.
. . That is, \(\displaystyle P(\text{at least one card}) \:=\:100\% \:=\:1\)

. . Hence, the numerator is: .\(\displaystyle (0.5)(1) \:=\:0.5\)


\(\displaystyle \text{Denominator: }\(\text{at least one card}) \:=\(V \cup M)\)

. . \(\displaystyle \text{We found this in part (A): }\(V \cup M) \:=\:0.65\)


\(\displaystyle \text{Therefore: }\( V\,|\text{ at least one card}) \:=\:\frac{0.5}{0.65} \;=\;\frac{10}{13}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

\(\displaystyle P(V' \cap M) \:=\(M) - P(V \cap M) \:=\:0.4 - 0.25 \:=\:0.15\)

\(\displaystyle P(V \cap M') \:=\(V) - P(V \cap M) \:=\:0.5 - 0.25 \:=\:0.25\)