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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

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 - \(10 * 1/(2^5)\)

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

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?

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?

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: 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

Bunuel please help me with this question. 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

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

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

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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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.

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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28 Oct 2014, 04:25

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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.

gmatclubot

Re: If the probability of rain on any given day is 50%, what is
[#permalink]
22 Jul 2015, 14:27

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