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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]

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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]

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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]

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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

$$P=(\frac{1}{2})^5=\frac{1}{32}$$.

$$P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4}$$.

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

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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: Veritas Combinatorics and Probability Question #12 [#permalink]

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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]

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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]

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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]

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27 Apr 2011, 20:40
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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]

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27 Apr 2011, 21:01
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wow Ian !! Thanks .... I didnt read "in a row"

3 consecutive rainy days -
SRRRS
SSRRR
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: tough probability question from veritas [#permalink]

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04 Oct 2013, 21:32
eeakkan wrote:
ıf the probability of rain on any given day is 50%, what is the probability that it will rain on at least 3 days ina row during a 5 days period?
a) 3/32 b) 1/4 c) 9/32 d) 5/16 e) 1/2

Because I have not understand the answer.

See.. the Possibility of Rain on any given day=1/2

Now lets take the 5 day period...

Now we want to make it rain for 3 continuous days so : 1.1.1.1/2.1/2 for rain can happen or not for last 2 days of the week
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Re: tough probability question from veritas [#permalink]

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04 Oct 2013, 22:14
i answered A , but realized did wrong interceprion's subtraction

what we need is (1/2*1/2*1/2)+(1/2*1/2*1/2)+(1/2*1/2*1/2) = 3/8, but we should subtract (1/2*1/2*1/2*1/2) as 4 days interception, so 3/8-1/16=5/16
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Re: tough probability question from veritas [#permalink]

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04 Oct 2013, 22:48
Name the days as 1, 2, 3, 4 and 5.
Given, probability of rain on any given day = 1/2 so probability of no rain on any day is also 1/2.

the question is finding probability for atleast 3 days rain.

we have 3 cases,

i) Exactly 3 days raining, that means days having rain are either 1,2,3 or 2,3,4 or 3,4,5. (Since 3 days in a row)

so probability = 3 * (1/2)^5

ii) Exactly 4 days raining, which means days having rain are 1,2,3,4 or 1,2,3,5 or 2,3,4,5 or 1,3,4,5

probability - 4* (1/2)^5

iii) Exactly 5 days raining, all 5 days raining

so prob. = (1/2)^5

Add all and we will get probability = 1/4. hence b.
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Re: Rain at least 3 days [#permalink]

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08 Oct 2013, 06:43
Bunuel wrote:
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

$$P=(\frac{1}{2})^5=\frac{1}{32}$$.

$$P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4}$$.

Hi Bunuel,

Why did you add up the probabilities of the 3 scenarios vs. get the product of the 3 (5/16)?
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Re: Rain at least 3 days [#permalink]

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08 Oct 2013, 06:49
Expert's post
pauc wrote:
Bunuel wrote:
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

$$P=(\frac{1}{2})^5=\frac{1}{32}$$.

$$P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4}$$.

Hi Bunuel,

Why did you add up the probabilities of the 3 scenarios vs. get the product of the 3 (5/16)?

Let me ask you a question: the probability of heads is 1/2, the probability of tails is also 1/2. What is the probability of getting a head OR a tail when flipping once? Is it 1/2+1/2=1 or 1/2*1/2=1/4?
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Re: Rain at least 3 days [#permalink]

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08 Oct 2013, 07:33
Bunuel wrote:
Let me ask you a question: the probability of heads is 1/2, the probability of tails is also 1/2. What is the probability of getting a head OR a tail when flipping once? Is it 1/2+1/2=1 or 1/2*1/2=1/4?

Won't it be just 1/2?
But I get what you mean when the probability problem is an OR/AND problem. I just get confused sometimes as to which of the 2 it is, especially if the problem doesn't explicitly state if it's an OR or an AND problem.
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Re: Rain at least 3 days [#permalink]

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08 Oct 2013, 07:36
Expert's post
pauc wrote:
Bunuel wrote:
Let me ask you a question: the probability of heads is 1/2, the probability of tails is also 1/2. What is the probability of getting a head OR a tail when flipping once? Is it 1/2+1/2=1 or 1/2*1/2=1/4?

Won't it be just 1/2?
But I get what you mean when the probability problem is an OR/AND problem. I just get confused sometimes as to which of the 2 it is, especially if the problem doesn't explicitly state if it's an OR or an AND problem.

The probability of getting either tails or heads is 100%. How else?
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Re: If the probability of rain on any given day is 50%, what is [#permalink]

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Re: If the probability of rain on any given day is 50%, what is [#permalink]

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22 Jul 2015, 14:27
Bunnuel...I understand your explanation. But I was wondering if there is a way to solve this using the C or P formula for each of the cases of 3, 4 or 5 in a row cases.
Re: If the probability of rain on any given day is 50%, what is   [#permalink] 22 Jul 2015, 14:27
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