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The probability of rain is on any given day next week.
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05 Aug 2018, 14:44
05 Aug 2018, 14:44
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Question Stats:
66% (00:57) correct
33% (00:59) wrong based on 160 sessions
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The probability of rain is \(\frac{1}{2}\) on any given day next week.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability that it rains on at least one of the 7 days next week |
\(\frac{127}{128}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The probability of rain is on any given day next week.
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07 Aug 2018, 05:09
07 Aug 2018, 05:09
3
The probability that it rains on one of 7 days = 1-probability that it does not rain on any of the 7 days.
probability for not raining on any of the 7 days = \((1/2)^7 = \frac{1}{2^7}\)
ans = \(1 - \frac{1}{2^7}\)
or, \(\frac{(2^7 - 1)}{2^7} = \frac{127}{128}\)
probability for not raining on any of the 7 days = \((1/2)^7 = \frac{1}{2^7}\)
ans = \(1 - \frac{1}{2^7}\)
or, \(\frac{(2^7 - 1)}{2^7} = \frac{127}{128}\)
Re: The probability of rain is on any given day next week.
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09 Aug 2018, 15:09
09 Aug 2018, 15:09
4
Expert Reply
Explanation
Since Quantity A is an “at least” problem, use the 1 – x shortcut. Rather than calculate the probability of rain on exactly 1 day next week, and then the probability of rain on exactly 2 days next week, and so on (after which you would still have to add all of the probabilities together!), instead calculate the probability of no rain at all on any day, and then subtract that number from 1.
That will give the combined probabilities for any scenarios that include rain on at least 1 day. Probability of NO rain for any of the 7 days =\(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{128}\)
Subtract this probability from 1: \(1-\frac{1}{128}=\frac{127}{128}\).
Quantities A and B are equal.
Since Quantity A is an “at least” problem, use the 1 – x shortcut. Rather than calculate the probability of rain on exactly 1 day next week, and then the probability of rain on exactly 2 days next week, and so on (after which you would still have to add all of the probabilities together!), instead calculate the probability of no rain at all on any day, and then subtract that number from 1.
That will give the combined probabilities for any scenarios that include rain on at least 1 day. Probability of NO rain for any of the 7 days =\(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{128}\)
Subtract this probability from 1: \(1-\frac{1}{128}=\frac{127}{128}\).
Quantities A and B are equal.
