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# A certain city has a chance of rain occurring on any given d

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Joined: 07 Jun 2014
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A certain city has a chance of rain occurring on any given d [#permalink]  30 Jul 2018, 11:04
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Question Stats:

66% (00:59) correct 33% (00:43) wrong based on 21 sessions
A certain city has a $$\frac{1}{3}$$ chance of rain occurring on any given day. In any given 3-day period, what is the probability that the city experiences rain?

(A) $$\frac{1}{3}$$
(B) $$\frac{8}{27}$$
(C) $$\frac{2}{3}$$
(D) $$\frac{19}{27}$$
(E) $$1$$
[Reveal] Spoiler: OA

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Joined: 07 Jan 2018
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Re: A certain city has a chance of rain occurring on any given d [#permalink]  31 Jul 2018, 07:06
Let us find the possibility of no rain on all of the three days.
It would be $$\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}$$

$$1- \frac{8}{27} = \frac{19}{27}$$

This gives possibility of rain at least on one of those 3 days.
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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1776 [0], given: 397

Re: A certain city has a chance of rain occurring on any given d [#permalink]  21 Aug 2018, 18:34
Expert's post
Explanation

In essence, the question is asking, “What is the probability that one or more days are rainy days?” since any single rainy day would mean the city experiences rain. In this case, employ the 1 – x shortcut, where the probability of rain on one or more days is equal to 1 minus the probability of no rain on any day.

Since the probability of rain is $$\frac{1}{3}$$ on any given day, the probability of no rain on any given day is $$1 - \frac{1}{3}=2/3$$ .

Therefore, the probability of no rain on three consecutive days is $$\frac{2}{3}\frac{2}{3}\frac{2}{3}=\frac{8}{27}$$.

Finally, subtract from 1 to find the probability that it rains on one or more days: P(1 or more days) = 1 – P(no rain) = $$1 - \frac{8}{27}=\frac{19}{27}$$.
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Re: A certain city has a chance of rain occurring on any given d   [#permalink] 21 Aug 2018, 18:34
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