Acheck of dorms revealed that 38% had refrigerators, 52% had tv’s 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 refrigerator but not both (c) neither a tv nor a refrigerator

a) 31%  b) 48% c) 31%

Explanation:

We have the probabilities:

P(had TV) = 0.52

P(had refrigerator)=0.38

P(had both a TV and a refrigerator)=0.21

a) a TV but no refrigerator

P(a TV but no refrigerator) =P(TV)- P(both a TV and a refrigerator) = 0.52 - 0.21 = 0.31

(b) a TV or refrigerator but not both

P(a TV or refrigerator but not both) = P(TV) + P(had refrigerator) - 2×P(had both a TV and a refrigerator) = 0.52+ 0.38 - (2×0.21) = 0.48

(c) neither a TV nor a refrigerator

P(neither a TV nor a refrigerator) = 1 - ( P(TV) + P(had refrigerator) - P(had both a TV and a refrigerator) ) = 1- (0.52 + 0.38 - 0.21) = 0.31

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answervisible;