Last visit was: 25 Apr 2024, 18:10 It is currently 25 Apr 2024, 18:10

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Manager
Manager
Joined: 20 Apr 2010
Posts: 154
Own Kudos [?]: 248 [19]
Given Kudos: 28
Concentration: Finacee, General Management
Schools:ISB, HEC, Said
 Q48  V28
Send PM
Most Helpful Reply
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4946
Own Kudos [?]: 7627 [12]
Given Kudos: 215
Location: India
Send PM
General Discussion
User avatar
Manager
Manager
Joined: 08 Nov 2010
Posts: 204
Own Kudos [?]: 496 [3]
Given Kudos: 161
 Q50  V41
GPA: 3.9
WE 1: Business Development
Send PM
User avatar
Senior Manager
Senior Manager
Joined: 01 Feb 2011
Posts: 309
Own Kudos [?]: 324 [1]
Given Kudos: 42
Send PM
Re: If the probability of rain on any given day in City X is 25%, what is [#permalink]
1
Kudos
this is a binomial distribution problem.

P(rain) = 1/4

getting rain on 3 days out of 4 = 4c3*((1/4)^3)*(3/4) = 3/64

Answer is C.
avatar
Manager
Manager
Joined: 20 Nov 2010
Posts: 88
Own Kudos [?]: 117 [1]
Given Kudos: 38
Send PM
Re: If the probability of rain on any given day in City X is 25%, what is [#permalink]
1
Kudos
3 days can be choosen out of 4 in 4C3 ways.
Probability of raining 3 days out of 4 = 4C3 * (1/4)^3 * (3/4) = 3 / 64
Board of Directors
Joined: 11 Jun 2011
Status:QA & VA Forum Moderator
Posts: 6072
Own Kudos [?]: 4689 [0]
Given Kudos: 463
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Send PM
Re: If the probability of rain on any given day in City X is 25%, what is [#permalink]
prashantbacchewar wrote:
If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

A. 1/4
B. 1/32
C. 3/64
D. 4/32
E. 3/4


\(P(r) = \frac{1}{4}\) & \(P(r') = \frac{3}{4}\)

So, Required Probability is \(= (\frac{1}{4})^3*\frac{3}{4}\)

Or, Required Probability is \(= \frac{3}{64}\)

Hence, answer will be \((C) \frac{3}{64}\)
Intern
Intern
Joined: 06 Nov 2016
Posts: 22
Own Kudos [?]: 21 [1]
Given Kudos: 2
Send PM
Re: If the probability of rain on any given day in City X is 25%, what is [#permalink]
1
Bookmarks
Abhishek009 wrote:
prashantbacchewar wrote:
If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

A. 1/4
B. 1/32
C. 3/64
D. 4/32
E. 3/4


\(P(r) = \frac{1}{4}\) & \(P(r') = \frac{3}{4}\)

So, Required Probability is \(= (\frac{1}{4})^3*\frac{3}{4}\)

Or, Required Probability is \(= \frac{3}{64}\)

Hence, answer will be \((C) \frac{3}{64}\)



Required Probability is \(= (\frac{1}{4})^3*\frac{3}{4}\) = 3/256?

You are missing a required combination value. 4c3 * (\frac{1}{4})^3*\frac{3}{4}[/m] = 3/64
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18761
Own Kudos [?]: 22054 [1]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: If the probability of rain on any given day in City X is 25%, what is [#permalink]
1
Bookmarks
Expert Reply
prashantbacchewar wrote:
If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

A. 1/4
B. 1/32
C. 3/64
D. 4/32
E. 3/4



We need to determine the probability of having 3 rainy days within a 4-day period. We are given that the probability of a rainy day is 1/4, and thus the probability of a non-rainy day is 3/4.

We can assume the first 3 days are rainy (R) and the last day is not rainy (N). Thus:

P(R-R-R-N) = 1/4 x 1/4 x 1/4 x 3/4 = 3/256

However, we need to determine in how many ways it can rain 3 out of 4 days. That number will be equivalent to how many ways we can arrange the letters R-R-R-N.

We use the indistinguishable permutations formula to determine the number of ways to arrange R-R-R-N: 4!/3! = 4 ways

Each of these 4 ways has the same probability of occurring. Thus, the total probability is:

4(3/256) = 12/526 = 3/64

Answer: C
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32680
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: If the probability of rain on any given day in City X is 25%, what is [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: If the probability of rain on any given day in City X is 25%, what is [#permalink]
Moderators:
Math Expert
92915 posts
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne