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The possible outcomes of a visit by three persons to an emergency room. Assuming that the visits are independent, with a probability of 0.646 of being an emergency, calculate the following:

The probability of two emergencies of of three arrivals.

The probability of no emergencies of of three arrivals.

The probability of at least two emergencies of of three arrivals.

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    $\begingroup$ What are your thoughts? $\endgroup$ – user99680 Mar 2 '14 at 7:33
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Hint:

Probability of an Emergency = p = 0.646, Probability of not an emergency = q = 1-.646 = 0.354

Three visits: n = 3, Use Binomial expression of $${n\choose r}.p^{r}.q^{(n-r)}$$.

$$P(r = 2) = {3\choose2}.p^{2}.q^{(3-2)}$$

Find out P(r=0) in the similar fashion

$$P(r>=2) = P(r=2)+P(r=3)$$

$${n\choose r} = \frac{n!}{r!(n-r)!}$$

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