Question

A check of dorm rooms on a large college campus revealed that 38% had refrigerators, 52% had tvs, and 21% had both a tv and a refrigerator. what's the probability that a randomly selected dorm room has
a.a tv but no refrigerator?
b.a tv or a refrigerator, but not both?
c.neither a tv nor a refrigerator?

Answer 1

A) 31%
B) 48%
C) 31%

Explanation
52% had TVs, including the 21% that had both.  52-21 = 31% of students that only had a TV.
38% had refrigerators, including the 21% that had both.  38-21=17% that had only a refrigerator; combining this with the 31% that had only a TV, 31+17 = 48%.
100-(17+21+31) = 100-69 = 31% had neither.

Answer 2

Using the percentages given, it is found that:

a) There is a 31% probability that a randomly selected dorm room has a tv but no refrigerator.

b) There is a 48% probability that a randomly selected dorm room has a tv or a refrigerator, but not both.

c) There is a 31% probability that a randomly selected dorm room has neither a tv nor a refrigerator.

Item a:

• 52% had tv's.
• Of those, 21% also had a refrigerator.

52 - 21 = 31

There is a 31% probability that a randomly selected dorm room has a tv but no refrigerator.

Item b:

• 31% have only a TV.
• 38 - 21 = 17% have only a refrigerator.

31 + 17 = 48

There is a 48% probability that a randomly selected dorm room has a tv or a refrigerator, but not both.

Item c:

The probability of at least one is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Hence:

$P(A \cup B) = 0.52 + 0.38 - 0.21 = 0.69$

1 - 0.69 = 0.31

There is a 31% probability that a randomly selected dorm room has neither a tv nor a refrigerator.

A similar problem is given at https://brainly.com/question/24707032