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

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If the probability of rain on any given day is 50%, what is [#permalink]  21 Aug 2010, 17:04
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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

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Re: Rain at least 3 days [#permalink]  21 Aug 2010, 20:33
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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}.

Answer: B.
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Re: Rain at least 3 days [#permalink]  21 Aug 2010, 17:19
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).
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Re: Rain at least 3 days [#permalink]  21 Aug 2010, 18:54
i also got the same answer , this is not the correct answer
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Re: Rain at least 3 days [#permalink]  21 Aug 2010, 20:45
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
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Re: Rain at least 3 days [#permalink]  22 Aug 2010, 09:23
damn! i also missed the same part
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Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 19:16
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!
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Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 19:25
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!
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Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 19:45
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?
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Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 20:34
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?
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Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 20:40
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.
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Re: Veritas Combinatorics and Probability Question #12 [#permalink]  27 Apr 2011, 21:01
wow Ian !! Thanks .... I didnt read "in a row"

3 consecutive rainy days -
SRRRS
SSRRR
RRRSS
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
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Re: Veritas Combinatorics and Probability Question #12 [#permalink]  28 Apr 2011, 20:17
VeritasPrepKarishma, as a Veritas Instructor, can you please answer this?

Thanks!
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Re: Rain at least 3 days [#permalink]  08 Aug 2011, 15:00
Just curious: is there an approach for handling this type of question when the two mutually exclusive outcomes have different probabilities? Say, for example, the probability of rain is only 40%.

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Re: Rain at least 3 days [#permalink]  23 Dec 2011, 03:44
I have a question
Most of the time, we can use conventional methods to solve probability questions, but in some situations we should count the favorite possibilities (like this question). How we can understand when we should count and when we should use normal ways?
Re: Rain at least 3 days   [#permalink] 23 Dec 2011, 03:44
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