BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Probability question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Mon May 26, 2008 4:12 pm
by ch0719 » Sun Oct 26, 2008 4:10 pm
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?

1/32
2/25
5/16
8/25
3/4
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 154
Joined: Tue Aug 26, 2008 12:59 pm
Location: Canada
Thanked: 4 times

by canuckclint » Sun Oct 26, 2008 5:01 pm
Let me edit my answer:

So you are looking for 3 rainy days out of 5.

rrrnn. This can be arranged (5 c 2) = 10 ways.

2 ^ 5 = 32 total possible combinations.

10/32 = 5/16
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members
Thanked: 331 times

by cramya » Sun Oct 26, 2008 8:39 pm
2 ^ 5 = 32 total possible combinations.


can u eloborate how u get the total combinations? I am missing something here. Thanks!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members
Thanked: 146 times

Re: Probability question

by parallel_chase » Sun Oct 26, 2008 11:07 pm
ch0719 wrote: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?

1/32
2/25
5/16
8/25
3/4
probability of rain on any given day = 50/100 = 1/2
probability of not raining on any given day = 1-1/2 = 1/2

no. of days = 4, 5, 6, 7, 8 = 5 days

exactly 3 days from July 4 - July 8

total combinations = 5C3 = 10

3 days raining * 2 days not raining = (1/2)^5

total probability = 10*(1/2)^5 = 10/2^5 = 5/16

Hope this helps.
No rest for the Wicked....
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Mon May 26, 2008 4:12 pm

by ch0719 » Sun Oct 26, 2008 11:07 pm
thanks, I over analyzed this question... :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 235
Joined: Wed Oct 26, 2016 9:21 pmFollowed by: 5 members
Thanked: 3 times

by Anaira Mitch » Thu Jan 05, 2017 6:57 pm
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.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3008
Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members
Location: Grand Central / New York
Thanked: 470 times

by Jay@ManhattanReview » Thu Jan 05, 2017 9:54 pm
ch0719 wrote: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?

1/32
2/25
5/16
8/25
3/4
We are given that the probability of rain on a given day = 1/2.

There are 5 days during July 4 through July 8, inclusive.

We wish that it rains on any of the three days and does not rain on the two days.

Probability that it does not rain on a given day = 1 - 1/2 = 1/2.

Say, it rains on the first three days and does not rain in the last days.

Probability that it rains on the first three days and does not rain in the last two days (RRRNN) = (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = (1/2)^5

Since the order of rainy days is not a constraint, i.e., it can rain on any three of the given five days, we have a combination problem.

Ways of choosing 3 days out of 5 days = 5C3 = 5C2 = (5.4)/(1.2) = 10 ways; [nCr = nC(n-r)]

Probability that it rains on any three days = 10*(1/2)^5 = 5/(2^4) = 5/16.

Answer: C

-Jay
_________________
Manhattan Review GMAT Prep

Locations: New York | Beijing | Auckland | Milan | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Fri Jan 06, 2017 10:20 am
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 consistent 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: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 1462
Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members
Location: New York, NY
Thanked: 39 times

by Jeff@TargetTestPrep » Mon Jan 09, 2017 9:47 am
ch0719 wrote: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?

1/32
2/25
5/16
8/25
3/4
The number of days from July 4 to July 8, inclusive, is 5 days. Thus, we need to determine the probability of having 3 rainy days within a 5-day period. We are given that the probability of a rainy day is ½, and thus the probability of no rain is also 1/2. We need to determine the probability of 3 rainy days within a 5-day period. We can assume the first 3 days are rainy (R) and the last 2 days are not rainy (N). Thus,

P(R-R-R-N-N) = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32

However, we need to determine in how many ways it can rain 3 out of 5 days. That number will be equivalent to how many ways we can organize the letters R-R-R-N-N.

We use the indistinguishable permutations formula to determine the number of ways to arrange R-R-R-N-N: 5!/(3! x 2!) = 10 ways

Each of these 10 ways has the same probability of occurring. Thus, the total probability is:

10(1/32) = 10/32 = 5/16

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1