Question

Consider randomly selecting a student at a certain 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) = 0.2, P(B) = 0.3, and P(A \cap B) = 0.05. (a) Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event A \cup B). (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. $\circ$ A' \cap B $\circ$ A \cap B' $\circ$ A \cup B' $\circ$ A' \cup B' $\circ$ A' \cap B' Calculate the probability of this event.

          Consider randomly selecting a student at a certain 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) = 0.2, P(B) = 0.3, and P(A \cap B) = 0.05.
(a) Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event A \cup B).
(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.
$\circ$ A' \cap B
$\circ$ A \cap B'
$\circ$ A \cup B'
$\circ$ A' \cup B'
$\circ$ A' \cap B'
Calculate the probability of this event.

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Consider randomly selecting a student at a certain 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) = 0.2, P(B) = 0.3, and P(A ∩B) = 0.05.
(a) Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event A ∪B).
(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.
∘ A' ∩B
∘ A ∩B'
∘ A ∪B'
∘ A' ∪B'
∘ A' ∩B'
Calculate the probability of this event.

Added by Ronald H.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Lucas Finney

02/13/2023

Step 1

Step 1: The probability of the event A U B is given by: P(A U B) = P(A) + P(B) - P(A ∩ B) Show more…

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Consider randomly selecting a student at a certain 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) = 0.2, P(B) = 0.3, and P(A ∩ B) = 0.05. (a) Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event A U B). (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. O A'N B Ο ΑΠΒ' AUB' OA'U B' Ο Α' Β' Calculate the probability of this event.

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00:01 All right, so for part a, the probability of a or b or c will be equal to p of a plus p of b plus p of c minus p of a and b minus p of a and c minus p of b and c minus p of a and c and c minus p of a and c and c.

00:32 C.

00:33 So plugging in our values given to us, we have p of a is 0 .6, p of b is 0 .4, p of, we're not given p.

00:42 Oh, actually, no, we are given p of c, pardon me.

00:45 That's 0 .2.

00:46 Then we do minus p of a and b, 0 .3, minus p of a and c, which is 0 .12.

00:53 Minus p of b and c, which is 0 .1, minus p of a, b, and c, which is 0 .08.

00:59 So that gives us a result of 0 .12.

01:01 So that gives us a result of 0 .0 .0 .0.

01:01 So that gives us a result of 0.

01:02 0 .6.

01:05 Then for part b, probability that, let's see here, both a visa and a mastercard, but not an american express.

01:17 So that's probability of a and b and c complement.

01:27 So the way that we would find that is, would be probability of a and b minus p of a and b and c.

01:44 So p of a and b is 0 .3, then we subtract off 0 .08 for a result of 0 .22.

01:51 Then moving on to part c, probability of b given a, so equal to probability of a and b divided by p of a, which would be equal to 0 .3 divided by 0 .6, which is 0 .5.

02:11 For part d, or pardon me, we're not all the way to part d yet...

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