Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 Aug 2015, 18:34

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If the probability of rain on any given day is 50%, what is

Author Message
TAGS:
Senior Manager
Joined: 16 Apr 2009
Posts: 340
Followers: 1

Kudos [?]: 79 [0], given: 14

If the probability of rain on any given day is 50%, what is [#permalink]  21 Aug 2010, 16:04
7
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

30% (03:14) correct 70% (01:14) wrong based on 261 sessions
If the probability of rain on any given day is 50%, what is the probability that it will rain on at least 3 days in a row during a 5 day period?

(A) 3/32
(B) 1/4
(C) 9/32
(D) 5/16
(E) 1/2
[Reveal] Spoiler: OA

_________________

Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 7

Kudos [?]: 142 [0], given: 50

Re: Rain at least 3 days [#permalink]  21 Aug 2010, 16:19
ichha148 wrote:
If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?
(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

Please explain . OA to be shared later

My attempt:

Rainy days - R, Non-Rainy days - NR.

Prob of at least 3 rainy days - 1 - (Prob of zero + Prob of 1 + Prob of 2 rainy days)

Prob of zero rainy days - NNNNN - $$1/(2^5)$$

Prob of 1 rainy day - NNNNR - $$1/(2^5)$$. But the one rainy day could be either first, second of third...

Hence total combination - $$5C1$$. Prob of 1 rainy day - $$5 * 1/(2^5)$$

Prob of 2 rainy day - NNNRR - $$1/(2^5)$$. But the one rainy day could be either first & second or third & fourth...

Hence total combination - 5C2. Prob of 1 rainy day - $$10 * 1/(2^5)$$

Hence total prob = 1 - ((1/32) + (5/32) + (10/32))

1- (16/32)

= 1/2. Answer should be 1/2 (E).
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Math Expert
Joined: 02 Sep 2009
Posts: 28784
Followers: 4606

Kudos [?]: 47696 [6] , given: 7130

Re: Rain at least 3 days [#permalink]  21 Aug 2010, 19:33
6
KUDOS
Expert's post
4
This post was
BOOKMARKED
ichha148 wrote:
If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?
(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

Please explain . OA to be shared later

It's basically the same as this one: hard-probability-99478.html?hilit=heads#p766970

At least 3 days in a row during a 5 day period means that 3, 4 or 5 days in a row:

3 days in a row
5 cases:
RRRNN
NRRRN
NNRRR
RNRRR
RRRNR

$$P=5*(\frac{1}{2})^5=\frac{5}{32}$$.

4 days in a row
2 cases:
RRRRN
NRRRR

$$P=2*(\frac{1}{2})^5=\frac{2}{32}$$.

5 days in a row
1 case:
RRRRR

$$P=(\frac{1}{2})^5=\frac{1}{32}$$.

$$P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4}$$.

_________________
Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 7

Kudos [?]: 142 [0], given: 50

If the probability of rain on any given day is 50%, what is [#permalink]  21 Aug 2010, 19:45
Thanks Bunuel.

I missed to capture the crucial wording of the question....

what is the probability that it will rain on at least 3 days [highlight]in a row[/highlight] during a 5 day period
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Senior Manager
Joined: 16 Apr 2009
Posts: 340
Followers: 1

Kudos [?]: 79 [0], given: 14

Re: Rain at least 3 days [#permalink]  22 Aug 2010, 08:23
damn! i also missed the same part
_________________

Intern
Joined: 27 Apr 2011
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 18:16
I am reviewing the Veritas Combinatorics and Probability book right now and have a question on question 12 (pg 54). It asks: If the probability of rain on any given day is 50%, what is the probability that it will rain on at least 3 days in a row during a 5 day period?

I understand the explanation given in the book. The method is a little tedious--it solves the problem by drawing out all the possibilities and then adding them together. Anyone know how to solve this by pure math/combinatorics instead of writing out all the possibilities?

Thanks!
Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 926
Followers: 13

Kudos [?]: 240 [0], given: 123

Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 18:25
Its like throwing a coin since the odds are equal. P(raining) = 0.5 and P(not raining) = 0.5 on any day

I think we can use the binomial distribution here. total probability = sum of probabilities (3 days rain + 4 days rain + 5 days rain)

P(3 days) = 5C3 (0.5)^3 (0.5)^2
P(4 days) = 5C4 (0.5)^4 (0.5)^1
P(5 days) = 5C5 (0.5)^5 (0.5)^0

P(atleast 3 days) = sum of above.

Conversely you can use the 1- x rule. P(at least 3 days ) = 1- P(atmost 2 days)

gmatstudier wrote:
I am reviewing the Veritas Combinatorics and Probability book right now and have a question on question 12 (pg 54). It asks: If the probability of rain on any given day is 50%, what is the probability that it will rain on at least 3 days in a row during a 5 day period?

I understand the explanation given in the book. The method is a little tedious--it solves the problem by drawing out all the possibilities and then adding them together. Anyone know how to solve this by pure math/combinatorics instead of writing out all the possibilities?

Thanks!
Intern
Joined: 27 Apr 2011
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 18:45
Yeah that's what I thought, but the answer is .25. I tried adding your equation up and it didn't match. Am I doing somehting wrong?
Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 926
Followers: 13

Kudos [?]: 240 [0], given: 123

Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 19:34
I find concurrence here. See Testluv approach -
http://www.beatthegmat.com/probability- ... 49765.html

Quote:
student22 wrote:
Thanks valleeny for clarifying.

I just came across a similar problem where instead of EXACTLY 3 out of 5 days rain
it says AT LEAST 3 out of 5 days rain. So you would have to calculate the chance of rain occurring on 3, 4, and 5 days:

You would do (5C3 + 5C4 + 5C5) / 2^5

16/32 = 1/2

Testluv, for the pascal's triangle in this situation you would go to the last row and add everything to the right of the 3rd digit 10 (Digit 4, 5, and 6)? so 10 + 5 + 1 = 16/32?
Yes, that's correct. (Alernatively, you can go to the last row, and add everything to the left of the fourth digit: 1 + 5 + 10 = 16.)

EDIT: Here's how we work with the last row. If asked for probability of 0 rainy days (or 0 heads, etc), then we look at either the "1" that begins the row or the "1" that ends the row. If we want to 1 rainy day, then we look at either of the "5"s. If we want 2 or 3 rainy days, then we look at either of the "10"s. If we want "at least" then we just sum up in the manner you intuited.

When a probability problem says "at least" it is almost always easier to think about the undesirable events, and then subtract from 1. So, using the formula, it would be easier to say: 1 - (5C0 + 5C1 + 5C2)/2^5 = 1/2. Here, it doesn't make a big difference but if the problem had said "at least 2 days of rain", it would be far quicker to compute 5C0 + 5C1, than it would be to compute 5C2 + 5C3 + 5C4 + 5C5.

Know that nCn = nC0 = 1. Also, nC1 = n. Knowing this greatly speeds up usage of the formula.

gmatstudier wrote:
Yeah that's what I thought, but the answer is .25. I tried adding your equation up and it didn't match. Am I doing somehting wrong?
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1174
Followers: 313

Kudos [?]: 985 [1] , given: 4

Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 19:40
1
KUDOS
Expert's post
It's not asking for the probability that it rains on any 3 (or more) days; it's asking for the probability it rains on (at least) 3 days *in a row*. That's going to be less likely than it just raining on 3 days, because we don't want to count some sequences, like Rain, Sun, Rain, Sun, Rain, where we don't have 3 rainy days in a row.

There are a lot of ways to count the sequences which have 3 Rs in a row. Let's think of a sequence of rainy and sunny days as a word containing R's and S's. We can count as follows:

* if our word starts with RRR, we'd have 2 choices for the fourth and fifth days, so 2*2 = 4 sequences which begin with three rainy days

* if our word starts with SRRR, we have 2 choices for the final day, for 2 sequences we haven't yet counted with three rainy days in the middle (we have already counted the possibilities where the first day is an R)

* if our word ends in RRR, we have 2 final possibilities we have not yet counted: RSRRR and SSRRR (we have counted the possibilities with three Rs in the middle, so we need to make sure only to count those possibilities where the second day is an S)

so, adding, there are a total of 8 sequences with at least three consecutive Rs, and since there are 2^5 = 32 total sequences, the probability of having at least three consecutive rainy days is 8/32 = 1/4.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 926
Followers: 13

Kudos [?]: 240 [0], given: 123

Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 20:01
wow Ian !! Thanks .... I didnt read "in a row"

3 consecutive rainy days -
SRRRS
SSRRR
RSRRR
RRRSR

4 consecutive rainy days -
RRRRS
SRRRR

5 consecutive rainy days -
RRRRR

Total ways = 8

P(atleast 3 consecutive days) = 8/2^5 = 8/32 = 0.25
Intern
Status: NOT READY TO GIVE UP
Joined: 24 Apr 2013
Posts: 28
Location: India
Concentration: Strategy, Marketing
GMAT Date: 10-30-2013
WE: Engineering (Other)
Followers: 0

Kudos [?]: 8 [0], given: 45

Re: tough probability question from veritas [#permalink]  04 Oct 2013, 20:32
eeakkan wrote:
ıf the probability of rain on any given day is 50%, what is the probability that it will rain on at least 3 days ina row during a 5 days period?
a) 3/32 b) 1/4 c) 9/32 d) 5/16 e) 1/2

Because I have not understand the answer.

See.. the Possibility of Rain on any given day=1/2

Now lets take the 5 day period...

Now we want to make it rain for 3 continuous days so : 1.1.1.1/2.1/2 for rain can happen or not for last 2 days of the week
Senior Manager
Joined: 23 Jan 2013
Posts: 347
Schools: Cambridge'16
Followers: 2

Kudos [?]: 27 [0], given: 33

Re: tough probability question from veritas [#permalink]  04 Oct 2013, 21:14
i answered A , but realized did wrong interceprion's subtraction

what we need is (1/2*1/2*1/2)+(1/2*1/2*1/2)+(1/2*1/2*1/2) = 3/8, but we should subtract (1/2*1/2*1/2*1/2) as 4 days interception, so 3/8-1/16=5/16
Intern
Joined: 28 Jan 2013
Posts: 34
Followers: 0

Kudos [?]: 15 [0], given: 20

Re: tough probability question from veritas [#permalink]  04 Oct 2013, 21:48
Name the days as 1, 2, 3, 4 and 5.
Given, probability of rain on any given day = 1/2 so probability of no rain on any day is also 1/2.

the question is finding probability for atleast 3 days rain.

we have 3 cases,

i) Exactly 3 days raining, that means days having rain are either 1,2,3 or 2,3,4 or 3,4,5. (Since 3 days in a row)

so probability = 3 * (1/2)^5

ii) Exactly 4 days raining, which means days having rain are 1,2,3,4 or 1,2,3,5 or 2,3,4,5 or 1,3,4,5

probability - 4* (1/2)^5

iii) Exactly 5 days raining, all 5 days raining

so prob. = (1/2)^5

Add all and we will get probability = 1/4. hence b.
Intern
Joined: 15 Sep 2013
Posts: 11
Followers: 0

Kudos [?]: 1 [0], given: 18

Re: Rain at least 3 days [#permalink]  08 Oct 2013, 05:43
Bunuel wrote:
ichha148 wrote:
If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?
(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

Please explain . OA to be shared later

It's basically the same as this one: hard-probability-99478.html?hilit=heads#p766970

At least 3 days in a row during a 5 day period means that 3, 4 or 5 days in a row:

3 days in a row
5 cases:
RRRNN
NRRRN
NNRRR
RNRRR
RRRNR

$$P=5*(\frac{1}{2})^5=\frac{5}{32}$$.

4 days in a row
2 cases:
RRRRN
NRRRR

$$P=2*(\frac{1}{2})^5=\frac{2}{32}$$.

5 days in a row
1 case:
RRRRR

$$P=(\frac{1}{2})^5=\frac{1}{32}$$.

$$P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4}$$.

Hi Bunuel,

Why did you add up the probabilities of the 3 scenarios vs. get the product of the 3 (5/16)?
Math Expert
Joined: 02 Sep 2009
Posts: 28784
Followers: 4606

Kudos [?]: 47696 [0], given: 7130

Re: Rain at least 3 days [#permalink]  08 Oct 2013, 05:49
Expert's post
pauc wrote:
Bunuel wrote:
ichha148 wrote:
If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?
(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

Please explain . OA to be shared later

It's basically the same as this one: hard-probability-99478.html?hilit=heads#p766970

At least 3 days in a row during a 5 day period means that 3, 4 or 5 days in a row:

3 days in a row
5 cases:
RRRNN
NRRRN
NNRRR
RNRRR
RRRNR

$$P=5*(\frac{1}{2})^5=\frac{5}{32}$$.

4 days in a row
2 cases:
RRRRN
NRRRR

$$P=2*(\frac{1}{2})^5=\frac{2}{32}$$.

5 days in a row
1 case:
RRRRR

$$P=(\frac{1}{2})^5=\frac{1}{32}$$.

$$P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4}$$.

Hi Bunuel,

Why did you add up the probabilities of the 3 scenarios vs. get the product of the 3 (5/16)?

Let me ask you a question: the probability of heads is 1/2, the probability of tails is also 1/2. What is the probability of getting a head OR a tail when flipping once? Is it 1/2+1/2=1 or 1/2*1/2=1/4?
_________________
Intern
Joined: 15 Sep 2013
Posts: 11
Followers: 0

Kudos [?]: 1 [0], given: 18

Re: Rain at least 3 days [#permalink]  08 Oct 2013, 06:33
Bunuel wrote:
Let me ask you a question: the probability of heads is 1/2, the probability of tails is also 1/2. What is the probability of getting a head OR a tail when flipping once? Is it 1/2+1/2=1 or 1/2*1/2=1/4?

Won't it be just 1/2?
But I get what you mean when the probability problem is an OR/AND problem. I just get confused sometimes as to which of the 2 it is, especially if the problem doesn't explicitly state if it's an OR or an AND problem.
Math Expert
Joined: 02 Sep 2009
Posts: 28784
Followers: 4606

Kudos [?]: 47696 [0], given: 7130

Re: Rain at least 3 days [#permalink]  08 Oct 2013, 06:36
Expert's post
pauc wrote:
Bunuel wrote:
Let me ask you a question: the probability of heads is 1/2, the probability of tails is also 1/2. What is the probability of getting a head OR a tail when flipping once? Is it 1/2+1/2=1 or 1/2*1/2=1/4?

Won't it be just 1/2?
But I get what you mean when the probability problem is an OR/AND problem. I just get confused sometimes as to which of the 2 it is, especially if the problem doesn't explicitly state if it's an OR or an AND problem.

The probability of getting either tails or heads is 100%. How else?
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 5736
Followers: 325

Kudos [?]: 64 [0], given: 0

Re: If the probability of rain on any given day is 50%, what is [#permalink]  28 Oct 2014, 03:25
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.
_________________
Intern
Joined: 16 Dec 2013
Posts: 21
Location: United States
GPA: 3.7
Followers: 0

Kudos [?]: 9 [0], given: 32

Re: If the probability of rain on any given day is 50%, what is [#permalink]  22 Jul 2015, 13:27
Bunnuel...I understand your explanation. But I was wondering if there is a way to solve this using the C or P formula for each of the cases of 3, 4 or 5 in a row cases.
Re: If the probability of rain on any given day is 50%, what is   [#permalink] 22 Jul 2015, 13:27
Similar topics Replies Last post
Similar
Topics:
6 If the probability of rain on any given day in Chicago during the summ 3 30 Oct 2014, 08:33
1 The chance of rain on any given day in Tel-Aviv is 50%. What 8 12 Nov 2012, 17:29
If the probability of rain on any given day in City X is 25%, what is 3 29 Sep 2011, 23:17
If the probability of rain on any given day in city x is 50% 4 12 Jul 2011, 01:23
10 If the probability of rain on any given day in City X is 50 15 04 Jan 2008, 15:01
Display posts from previous: Sort by