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Complementary Probability

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Complementary Probability

by knight247 » Fri Sep 09, 2011 3:54 am
If the probability of rain on any given day is 50%, what is the probability that it will rain on atleast 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

Problem is pretty straight forward. The total number of possible outcomes is 32
Our required outcomes are
RRRNN which can be arranged in 3 ways
RRRRN which can be arranged in 6-2=4 ways(Consider RRR=X So we have XRN which can be arranged in 3! ways but XRN=RXN and NRX=NXR so we subtract the 2 from 3!)
RRRRR which can only be arranged in 1 way

So total of 8 favourable outcomes. 8/32=1/4 Hence B

Ok, now here is where I have an issue. While solving math problems I don't just solve with the intent of getting the right answer but rather to understand the concept inside-out. So I was trying to solve this problem using the complementary method.

Probability of rain on atleast 3 consecutive days=1-Probability of rain on a maximum of 2 consecutive days

Hope that is correct. And hoping someone can help me solve using the complementary method. May sound silly, I know but would appreciate any help I can get. Thanks a ton.
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by cans » Fri Sep 09, 2011 4:17 am
atleast 3 means: exactly 3 + exactly 4 + exactly 5.
3: 5C3*(1/2)^3*(1/2)^2
4: 5C4*(1/2)^4*(1/2)
5: 5C5*(1/2)^5
total = (1/2)^5 * 16 = 1/2
IMO E

rrrnn can be arranged in 10 ways: 5!/(3!*2!)
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by knight247 » Fri Sep 09, 2011 4:33 am
@cans
Atleast 3 in a row is what the question asks...Not the same as atleast 3...Example we need RRRNN or NRRRN but not RRNNR.
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by GMATGuruNY » Fri Sep 09, 2011 4:43 am
knight247 wrote:If the probability of rain on any given day is 50%, what is the probability that it will rain on atleast 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

Problem is pretty straight forward. The total number of possible outcomes is 32
Our required outcomes are
RRRNN which can be arranged in 3 ways
RRRRN which can be arranged in 6-2=4 ways(Consider RRR=X So we have XRN which can be arranged in 3! ways but XRN=RXN and NRX=NXR so we subtract the 2 from 3!)
RRRRR which can only be arranged in 1 way

So total of 8 favourable outcomes. 8/32=1/4 Hence B

Ok, now here is where I have an issue. While solving math problems I don't just solve with the intent of getting the right answer but rather to understand the concept inside-out. So I was trying to solve this problem using the complementary method.

Probability of rain on atleast 3 consecutive days=1-Probability of rain on a maximum of 2 consecutive days

Hope that is correct. And hoping someone can help me solve using the complementary method. May sound silly, I know but would appreciate any help I can get. Thanks a ton.
I posted a solution for this problem here:

https://www.beatthegmat.com/rain-check-t79099.html

To count the number of ways NOT to get at least 3 days of rain in a row:

Let N = no rain and R= rain.
There are 5 days, with 2 possible outcomes each day.
Total possible outcomes = 2*2*2*2*2 = 32.

Number of ways to get no rain:
NNNNN = 1 way.

Number of ways to get 1 day of rain:
Number of ways to arrange NNNNR = 5!/4! = 5.

Number of ways to get 2 days of rain:
Number of ways to arrange NNNRR = 5!/(3!2!) = 10.

Number of ways to get 3 days of rain NOT in a row:
Number of ways to arrange NNRRR = 5!(2!3!) = 10.
We must subtract the arrangements in which there are 3 days of rain in a row: NNRRR, NRRRN, RRRNN = 3 ways.
Total ways = 10-3 = 7.

Number of ways to get 4 days of rain but no more than 2 in a row:
RRNRR = 1 way.

Total number of ways NOT to get at least 3 days of rain in a row:
1+5+10+7+1 = 24.

Thus, the total number of ways to GET at least 3 days of rain in a row = 32-24 = 8.

P(at least 3 days of rain in a row) = 8/32 = 1/4.
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