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3 years ago

5

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 rando

m room does not contain disease y?

2 answers:

Zolol [24]3 years ago

6 0

Answer:

6/10

Step-by-step explanation:

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Anna71 [15]3 years ago

3 0

Answer:

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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Given:
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Step-by-step explanation:

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3 years ago

5m - 4n, when m=5 and n = 3.

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Answer:

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2 years ago

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Juli2301 [7.4K]

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Step-by-step explanation:

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2 years ago

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Alex17521 [72]

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