A check of dorm rooms on a large college campus revealed that 34 % had refrigerators, 51 % had TVs, and 21 % had both a refrigerator and a TV. What's the probability that a randomly selected dorm room has
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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 hasa.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% = 0.21a. Probability of a TV but no refrigerator = P(A∩B')= P(A) - P(A∩B)= 0.34 - 0.21= 0.31b. 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)= 0.43c. 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