Question

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?

Answer 1

6/10

Explanation:

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Answer 2

6/10

Explanation:

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