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# PS: MGMAT Challenge of the week

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VP
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PS: MGMAT Challenge of the week [#permalink]

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03 Nov 2005, 22:55
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

All gnomes that solve this problem will get some underpants from our friends at Manhattan on Monday, when they provide OA/OE for this Try it out.

In October 2005, there was a record 75% chance of rain in New York City on any given day (along with a 25% chance of no rain). Assuming that each dayâ€™s weather was independent of the weather on any other day, what was the probability that in a given week, it rained at least 5 days in a row?

(A) (3/4)^5 + (3/4)^6 + (3/4)^7

(B) (3/4)^5 (1/4)^2 + (3/4)^6 (1/4) + (3/4)^7

(C) 3 * (3/4)^5 (1/4)^2 + 4 * (3/4)^6 (1/4) + (3/4)^7

(D) (7!/5!) * (3/4)^5 (1/4)^2 + (7!/6!) * (3/4)^6 (1/4) + (3/4)^7

(E) (7!/(5! * 2!)) * (3/4)^5 (1/4)^2 + (7!/6!) * (3/4)^6 (1/4) + (3/4)^7

Last edited by duttsit on 04 Nov 2005, 16:20, edited 1 time in total.

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Director
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Re: PS: MGMAT Challenge of the week [#permalink]

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03 Nov 2005, 23:13
duttsit wrote:
All gmones that solve this problem will get some underpants from our friends at Manhattan on Monday, when they provide OA/OE for this Try it out.

In October 2005, there was a record 75% chance of rain in New York City on any given day (along with a 25% chance of no rain). Assuming that each dayâ€™s weather was independent of the weather on any other day, what was the probability that in a given week, it rained at least 5 days in a row?

This is what I get so far using the binomial theorem:

rain 5 days in a row

7C5*(3/4)^5*(1/4)^2

rain 6 days in a row

7C6*(3/4)^6*(1/4)^1

rain 7 days in a row

7C7*(3/4)^7*(1/4)^0

Add em up and you get

(3/4)^5*(1/4)^2*(7!/5!2!)+(3/4)^6*(1/4)*(7!/6!)+(3/4)^7

which is E in MGMAT's Answer choice

do i get underpants just for trying???

what did you get duttsit?
_________________

"Wow! Brazil is big." â€”George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

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VP
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03 Nov 2005, 23:24
Titleist, you will get underpants if I do. My answer/strategy is same as yours

anyone else?

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Director
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03 Nov 2005, 23:27
duttsit wrote:
Titleist, you will get underpants if I do. My answer/strategy is same as yours

anyone else?

Really??? You made my day or night I should say! Darn, it's already 1:26am my time - gots to get some sleep. Goodnight Duttsit! I'll dream of getting more underpants!
_________________

"Wow! Brazil is big." â€”George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

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Current Student
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04 Nov 2005, 10:45
OK

probability of raining 5 days in a row...welll

(0.75)^5 ?

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Director
Joined: 24 Oct 2005
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04 Nov 2005, 11:48
I know how you guys approached the problem.
But I was wondering if one of you could help me with my probability concept here..

When they say 5 days in a row.. It means. MTWHF OR TWHFS or WHFSS..
It does not mean MTWHS.. which is 5 days in a week..
So, 7C5 GIVES US.. number of ways I can choose 5 numbers out of 7, right? so how does that relate to the question..
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VP
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Location: CA
Re: PS: MGMAT Challenge of the week [#permalink]

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04 Nov 2005, 12:55
Titleist wrote:
This is what I get so far using the binomial theorem:

rain 5 days in a row

7C5*(3/4)^5*(1/4)^2

rain 6 days in a row

7C6*(3/4)^6*(1/4)^1

rain 7 days in a row

7C7*(3/4)^7*(1/4)^0

Add em up and you get

(3/4)^5*(1/4)^2*(7!/5!2!)+(3/4)^6*(1/4)*(7!/6!)+(3/4)^7

which is E in MGMAT's Answer choice

do i get underpants just for trying???

what did you get duttsit?

Looks like we need to relook at this question. There is more than what meets the eyes.

First of all, using theorem, we can find that:

p(rain) = 3/4
p(no rain) = 1/4

we need to find:
p(rain for at least 5 days in a row)

this is equivalent to:
p(raining 5 days in a row) + p(raining 6 days in a row, 5 of which should be in a row) + p(raining for all 7 days)

note: second component, it is not 6 days in a row

R: Rain
N: No rain

lets find invidual components:
p(raining 5 days in a row) =

RRRRRNN
NRRRRRN
NNRRRRR

each of these have probability: (3/4)^5 (1/4)^2
so p(raining 5 days in a row) = 3 * (3/4)^5 (1/4)^2

p(raining 6 days in a row, 5 of which should be in a row) =
RRRRRRN
NRRRRRR
RRRRRNR
RNRRRRR

each of these have probability (3/4)^6 * (1/4)^1
p(raining 6 days in a row, 5 of which should be in a row) =
4 * (3/4)^6 * (1/4)

p(raining for all 7 days) = (3/4)^7 simple

plugging all these components back gives:
p(raining atleast 5 days in a row) =

3 * (3/4)^5 (1/4)^2 + 4 * (3/4)^6 * (1/4)
+ (3/4)^7

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Director
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04 Nov 2005, 13:00
bewakoof wrote:
I know how you guys approached the problem.
But I was wondering if one of you could help me with my probability concept here..

When they say 5 days in a row.. It means. MTWHF OR TWHFS or WHFSS..
It does not mean MTWHS.. which is 5 days in a week..
So, 7C5 GIVES US.. number of ways I can choose 5 numbers out of 7, right? so how does that relate to the question..

I'm giving it a quick relook - and I think I need to adjust the number of days that it can rain 5 days in a row to 3 choices

RRRRRNN
NNRRRR
NRRRRN

So only 3 choices x (3/4)^5x(1/4)^2

For at least 5 or 6 days you can have 4 choices

NRRRRRR
RRRRRRN
RNRRRRR
RRRRRNR

So 4 choices x (3/4)^5x(1/4)^2

And of course there's only 1 way it can rain 7 days in a row

thus (3/4)^7

So it is Answer choice C

I won't bother doing MGMAT problems at 2am anymore.
_________________

"Wow! Brazil is big." â€”George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

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Director
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04 Nov 2005, 13:11
Bewakoof thanks for pointing that out!
_________________

"Wow! Brazil is big." â€”George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

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Director
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Re: PS: MGMAT Challenge of the week [#permalink]

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04 Nov 2005, 15:18
Probabilty of rain on at least 5 days in row

(3/7C5)*(3/4)^5*(1/4)^2+(2/7C6)*(3/4)^6*(1/4)^1+(1/7C7)*(3/4)^7

which is 0.1483154

RK

duttsit wrote:
All gmones that solve this problem will get some underpants from our friends at Manhattan on Monday, when they provide OA/OE for this Try it out.

In October 2005, there was a record 75% chance of rain in New York City on any given day (along with a 25% chance of no rain). Assuming that each dayâ€™s weather was independent of the weather on any other day, what was the probability that in a given week, it rained at least 5 days in a row?

Kudos [?]: 176 [0], given: 0

Director
Joined: 14 Oct 2003
Posts: 582

Kudos [?]: 59 [0], given: 0

Location: On Vacation at My Crawford, Texas Ranch

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07 Nov 2005, 15:16
OA IS C!

The probability that we are being asked to calculate here is comprised of multiple scenarios. Put differently, there are a number of ways for it to rain at least 5 days in a row. To solve for the overall probability, we must find the probability for each one of these independent scenarios and add them together.

Letâ€™s set: R = rainy day
S = sunny day

There are 3 different types of scenarios that we need to consider: (1) 5Râ€™s 2Sâ€™s (2) 6Râ€™s 1S and (3) 7Sâ€™s. Of course we must remember that within each scenario it must rain for at least 5 consecutive days.

TYPE 1: There are 3 ways for it to rain exactly 5 out of 7 days and for the 5 days to be consecutive: RRRRRSS, SSRRRRR, SRRRRRS (two Sâ€™s at the end, the beginning, or one on each end).

TYPE 2: There are 4 ways for it to rain exactly 6 out of 7 days and for at least five of the rainy days to be consecutive: RRRRRRS, SRRRRRR, RSRRRRR, RRRRRSR (S in position 1, 2, 6 or 7).

TYPE 3: There is one way for it to rain exactly 7 out of 7 days: RRRRRRR.

Now that we have counted and identified the different scenarios, we must find the probability of each scenario (or at least of each type of scenario).

TYPE 1: If the probability of R is Â¾ and the probability of S is Â¼, the probability of 5Râ€™s and 2Sâ€™s (it doesnâ€™t matter what order) is (3/4)^5(1/4)^2. There are three such scenarios so this contributes 3(3/4)^5(1/4)^2 to the final probability.

TYPE 2: If the probability of R is Â¾ and the probability of S is Â¼, the probability of 6Râ€™s and 1S (it doesnâ€™t matter what order) is (3/4)^6(1/4)^1. There are four such scenarios so this contributes 4(3/4)^6(1/4) to the final probability.

TYPE 3: If the probability of R is Â¾ and the probability of S is Â¼, the probability of 7Râ€™s is (3/4)^7. There is only one such scenario so this contributes (3/4)^7 to the final probability.

The final probability is 3(3/4)^5(1/4)^2 + 4(3/4)^6(1/4) + (3/4)^7.

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"Wow! Brazil is big." â€”George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

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VP
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07 Nov 2005, 15:23
Wow..Titleist. you got it right. You will surely get some underpants from Manhattan guys

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Director
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07 Nov 2005, 15:25
duttsit wrote:
Wow..Titleist. you got it right. You will surely get some underpants from Manhattan guys

I only accept used underpants...
_________________

"Wow! Brazil is big." â€”George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

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07 Nov 2005, 15:25
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