PS: MGMAT Challenge of the week

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!

OK

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

(0.75)^5 ?

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..

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

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.

Bewakoof thanks for pointing that out!

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

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.

Let’s set: R = rainy day
S = sunny day

There are 3 different types of scenarios that we need to consider: (1) 5R’s 2S’s (2) 6R’s 1S and (3) 7S’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 S’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 ¾ and the probability of S is ¼, the probability of 5R’s and 2S’s (it doesn’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 ¾ and the probability of S is ¼, the probability of 6R’s and 1S (it doesn’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 ¾ and the probability of S is ¼, the probability of 7R’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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Wow..Titleist. you got it right. You will surely get some underpants from Manhattan guys

I only accept used underpants...