PS: MGMAT Challenge of the week

Question

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

Titleist · Answer

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??? :rotate

what did you get duttsit?

what did you get duttsit?

duttsit · Answer

Titleist, you will get underpants if I do. My answer/strategy is same as yours

anyone else?

Titleist · Answer

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! :sleep

FN · Answer

OK

probability of raining 5 days in a row...welll

(0.75)^5 ?

bewakoof · Answer

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..

duttsit · Answer

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

Titleist · Answer

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 Thus the answer is 3(3/4)^5*(1/4)^2+4(3/4)^6(1/4)+(3/4)^7 So it is Answer choice C I won't bother doing MGMAT problems at 2am anymore. :oops:

Titleist · Answer

Bewakoof thanks for pointing that out! :cool

cool_jonny009 · Answer

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 8-)

RK

Titleist · Answer

OA IS C! 

Answer
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.
 
The correct answer is C. 
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duttsit · Answer

Wow..Titleist. you got it right. You will surely get some underpants from Manhattan guys

Titleist · Answer

I only accept used underpants...

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