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The probability of rain is on any given day next week.

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The probability of rain is on any given day next week. [#permalink] New post 05 Aug 2018, 14:44
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Question Stats:

64% (01:00) correct 35% (00:47) wrong based on 98 sessions
The probability of rain is \(\frac{1}{2}\) on any given day next week.

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.
[Reveal] Spoiler: OA

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Re: The probability of rain is on any given day next week. [#permalink] New post 07 Aug 2018, 05:09
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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, \(\frac{(2^7 - 1)}{2^7} = \frac{127}{128}\)
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Re: The probability of rain is on any given day next week. [#permalink] New post 09 Aug 2018, 15:09
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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.
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Re: The probability of rain is on any given day next week. [#permalink] New post 03 Jan 2019, 17:37
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Answer: D
Set = {x+1, x+2, x+3, …., x+35}

A: The probability for selecting an even number less than median:
The median is (35-1)/2+1 = 18th which is x+18
Numbers before median are: x+1, x+2, …, x+17 there are 17 numbers before median, and as 17 is not even, we don’t know there are either 9 or 8 even numbers.
A would be either 9/35 or 8/35

B: The probability for selecting an odd number greater than median:
The median is (35-1)/2+1 = 18th which is x+18
Numbers after median are: x+19, x+19, x+20, …, x+35, there are 17 numbers after median, we don’t know there are either 9 or 8 even numbers.
B would be either 9/35 or 8/35

We don’t know the first number in the sequence is even or odd, if we knew we were sure what situation would happen.
Answer is D
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Re: The probability of rain is on any given day next week. [#permalink] New post 18 Jul 2019, 03:58
sandy wrote:
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.



It is said that the probability of raining at any given day next week is 1/2. SO why you took 1/2 for no raining day. I'm confused here. please clarify.
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Re: The probability of rain is on any given day next week. [#permalink] New post 18 Jul 2019, 06:58
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huda wrote:


It is said that the probability of raining at any given day next week is 1/2. SO why you took 1/2 for no raining day. I'm confused here. please clarify.


If, for example, P(rain in Monday) = 0.5, then P(NO rain on Monday) = 1 - 0.5 = 0.5

Likewise, if it were the case that P(rain in Monday) = 0.3, then P(NO rain on Monday) = 1 - 0.3 = 0.7, etc.

Does that help?

Cheers,
Brent
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Re: The probability of rain is on any given day next week. [#permalink] New post 18 Jul 2019, 07:27
GreenlightTestPrep wrote:
huda wrote:


It is said that the probability of raining at any given day next week is 1/2. SO why you took 1/2 for no raining day. I'm confused here. please clarify.


If, for example, P(rain in Monday) = 0.5, then P(NO rain on Monday) = 1 - 0.5 = 0.5

Likewise, if it were the case that P(rain in Monday) = 0.3, then P(NO rain on Monday) = 1 - 0.3 = 0.7, etc.

Does that help?

Cheers,
Brent

yes, thank you so much. :yes :thanks :thanks :done
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Re: The probability of rain is on any given day next week.   [#permalink] 18 Jul 2019, 07:27
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