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# If the probability of rain on any given day in Chicago during the summ

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Math Expert
Joined: 02 Sep 2009
Posts: 46167
If the probability of rain on any given day in Chicago during the summ [#permalink]

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30 Oct 2014, 09:33
14
00:00

Difficulty:

45% (medium)

Question Stats:

69% (01:23) correct 31% (01:33) wrong based on 306 sessions

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Tough and Tricky questions: Combinations.

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

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Re: If the probability of rain on any given day in Chicago during the summ [#permalink]

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30 Oct 2014, 17:48
2
2
Bunuel wrote:

Tough and Tricky questions: Combinations.

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

From July 4 to July 8, we have total of 5 days. Prob of raining on exactly 3 days = $$5C3*(1/2)^3*(1/2)^2$$

= 5/16

C
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Re: If the probability of rain on any given day in Chicago during the summ [#permalink]

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01 Nov 2014, 08:20
Thoughtosphere wrote:
Bunuel wrote:

Tough and Tricky questions: Combinations.

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

From July 4 to July 8, we have total of 5 days. Prob of raining on exactly 3 days = $$5C3*(1/2)^3*(1/2)^2$$

= 5/16

C

Could you elaborate a little on you reasoning, I understand the $$5C3$$ as the number of combination of three days that we can have from 5 given days, I'm assuming wouldn't that be the denominator? ...

Thanks!
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Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: If the probability of rain on any given day in Chicago during the summ [#permalink]

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18 May 2015, 07:31
2
4
Bunuel wrote:

Tough and Tricky questions: Combinations.

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

Responding to a pm:

There are 5 days from July 4 to July 8 inclusive. You need the probability of 3 rainy and 2 non rainy days.

Method 1:
Probability = Favorable outcomes/Total outcomes

Total outcomes = 2*2*2*2*2 = 32
On each day, two things are possible - either it is rainy or non rainy. So the 5 days can happen in 2^5 = 32 ways e.g.
RRNNN, RNNNN, NNRNR, RRRRR etc

Favorable outcomes - 3 R days and 2 N days e.g.
RRRNN, RNRNR, RRNNR etc
In how many ways can you arrange 3 Rs and 2 Ns? 5!/3!*2! = 10

Probability = 10/32 = 5/16

Method 2:
Probability of rain = 1/2
Probability of no rain = 1/2

Probability of RRRNN = (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1/32
But there are other combinations too such as RRNRN, NNRRR etc. There are 10 such combinations as calculated above (in bold).
So total probability of 3 Rs and 2 Ns = (1/32) * 10 = 5/16
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 26 Oct 2016 Posts: 666 Location: United States Concentration: Marketing, International Business Schools: HBS '19 GMAT 1: 770 Q51 V44 GPA: 4 WE: Education (Education) Re: If the probability of rain on any given day in Chicago during the summ [#permalink] ### Show Tags 05 Jan 2017, 19:57 The period from July 4 to July 8, inclusive, contains 8 – 4 + 1 = 5 days, so we can rephrase the question as “What is the probability of having exactly 3 rainy days out of 5?” Since there are 2 possible outcomes for each day (R = rain or S = shine) and 5 days total, there are 2 x 2 x 2 x 2 x 2 = 32 possible scenarios for the 5 day period (RRRSS, RSRSS, SSRRR, etc…) To find the probability of having exactly three rainy days out of five, we must find the total number of scenarios containing exactly 3 R’s and 2 S’s, that is the number of possible RRRSS anagrams: = 5! / 2!3! = (5 x 4)/2 x 1 = 10 The probability then of having exactly 3 rainy days out of five is 10/32 or 5/16. _________________ Thanks & Regards, Anaira Mitch EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11801 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If the probability of rain on any given day in Chicago during the summ [#permalink] ### Show Tags 20 Jan 2018, 16:14 Hi All, The most efficient way to answer this question is in understanding how the Combination Formula applies to the math involved. However, even if you don't fully understand that concept, you can still get to the correct answer with a bit of 'brute force.' To start, you have to understand that since each day has two equally-likely outcomes ('rain' or 'not rain'), then the various arrangements of 5 days of weather = (2)(2)(2)(2)(2) = 32. Thus, the probability of having EXACTLY 3 rain days must be some fraction out of 32... From this point, we just have to figure out how many options consist of just 3 rain days. We can 'map' those out rather easily... RRRNN RRNRN RRNNR RNRRN RNRNR RNNRR NRRRN NRRNR NRNRR NNRRR 10 total options out of 32. Reducing that fraction gives us... 10/32 = 5/16 Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: If the probability of rain on any given day in Chicago during the summ [#permalink]

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28 May 2018, 07:57
VeritasPrepKarishma wrote:
Bunuel wrote:

Tough and Tricky questions: Combinations.

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

Responding to a pm:

There are 5 days from July 4 to July 8 inclusive. You need the probability of 3 rainy and 2 non rainy days.

Method 1:
Probability = Favorable outcomes/Total outcomes

Total outcomes = 2*2*2*2*2 = 32
On each day, two things are possible - either it is rainy or non rainy. So the 5 days can happen in 2^5 = 32 ways e.g.
RRNNN, RNNNN, NNRNR, RRRRR etc

Favorable outcomes - 3 R days and 2 N days e.g.
RRRNN, RNRNR, RRNNR etc
In how many ways can you arrange 3 Rs and 2 Ns? 5!/3!*2! = 10

Probability = 10/32 = 5/16

Method 2:
Probability of rain = 1/2
Probability of no rain = 1/2

Probability of RRRNN = (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1/32
But there are other combinations too such as RRNRN, NNRRR etc. There are 10 such combinations as calculated above (in bold).
So total probability of 3 Rs and 2 Ns = (1/32) * 10 = 5/16

Hi VeritasPrepKarishma, Bunuel

For someone who is particularly weak at Probability and Combinatorics, where would you recommend practice for ground work?
I have read through the theory but the application for under 700 Questions is what I need to work on.

thanks in anticipation.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: If the probability of rain on any given day in Chicago during the summ [#permalink]

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07 Jun 2018, 05:54
VeritasPrepKarishma wrote:
Bunuel wrote:

Tough and Tricky questions: Combinations.

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

Responding to a pm:

There are 5 days from July 4 to July 8 inclusive. You need the probability of 3 rainy and 2 non rainy days.

Method 1:
Probability = Favorable outcomes/Total outcomes

Total outcomes = 2*2*2*2*2 = 32
On each day, two things are possible - either it is rainy or non rainy. So the 5 days can happen in 2^5 = 32 ways e.g.
RRNNN, RNNNN, NNRNR, RRRRR etc

Favorable outcomes - 3 R days and 2 N days e.g.
RRRNN, RNRNR, RRNNR etc
In how many ways can you arrange 3 Rs and 2 Ns? 5!/3!*2! = 10

Probability = 10/32 = 5/16

Method 2:
Probability of rain = 1/2
Probability of no rain = 1/2

Probability of RRRNN = (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1/32
But there are other combinations too such as RRNRN, NNRRR etc. There are 10 such combinations as calculated above (in bold).
So total probability of 3 Rs and 2 Ns = (1/32) * 10 = 5/16

Hi VeritasPrepKarishma, Bunuel

For someone who is particularly weak at Probability and Combinatorics, where would you recommend practice for ground work?
I have read through the theory but the application for under 700 Questions is what I need to work on.

thanks in anticipation.

You can search for relevant tags (Probability, Below 600 and 600 - 700 level) on this forum itself.
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Re: If the probability of rain on any given day in Chicago during the summ   [#permalink] 07 Jun 2018, 05:54
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