Answer:
Consider the complete question is,
"A check of dorm rooms on a large college campus revealed that 34% had refrigerators, 51% 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?"
Solution : Suppose A = event of having refrigerators,
B = event of having a TV,
We have,
P(A) = 34% = 0.34,
P(B) = 51% = 0.51,
P(A∩B) = 21% = 0.21
a. Probability of a TV but no refrigerator = P(A∩B')
= P(A) - P(A∩B)
= 0.34 - 0.21
= 0.31
b. Probability of a TV or a refrigerator, but not both = P(A∪B) - P(A∩B)
= P(A) + P(B) - P(A∩B) - P(A∩B)
= P(A) + P(B) - 2P(A∩B)
= 0.34 + 0.51 - 2(0.21)
= 0.43
c. Probability of neither a TV nor a refrigerator = P(A' ∩ B')
= 1 - P(A∪B)
= 1 - P(A) - P(B) + P(A∩B)
= 1 - 0.34 - 0.51 + 0.21
= 0.36
