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27 November, 22:58

3 rooms are occupied by disease x, 4 by disease y, 2 rooms by disease z and 1 by disease a. What is the probability that a random room does not contain disease y?

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  1. 27 November, 23:32
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    6/10

    Step-by-step explanation:

    [P (x) + P (z) + P (a) ]/P (all) = (3+2+1) / (3+4+2+1) = 6/10
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