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24 February, 06:26

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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  1. 24 February, 06:41
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    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
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