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A certain city has a chance of rain occurring on any given d
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30 Jul 2018, 11:04
30 Jul 2018, 11:04
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79% (01:04) correct
20% (00:58) wrong based on 86 sessions
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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) \(1\)
(A) \(\frac{1}{3}\)
(B) \(\frac{8}{27}\)
(C) \(\frac{2}{3}\).
(D) \(\frac{19}{27}\)
(E) \(1\)
Re: A certain city has a chance of rain occurring on any given d
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31 Jul 2018, 07:06
31 Jul 2018, 07:06
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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.
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.
Re: A certain city has a chance of rain occurring on any given d
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21 Aug 2018, 18:34
21 Aug 2018, 18:34
1
Expert Reply
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) = \(1 - \frac{8}{27}=\frac{19}{27}\).
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) = \(1 - \frac{8}{27}=\frac{19}{27}\).
