Probability Problem

**mathhomework** · January 31st 2009, 04:22 PM

In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.

**ThePerfectHacker** · January 31st 2009, 04:39 PM

Originally Posted by mathhomework

In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.

Form a list of 7 empty spaces:
_ _ _ _ _ _ _

If it rains on Sunday, Monday, Tuesday, Wednesday, and Friday you write:
X X X X X _ _

There are ${7\choose 5} = {7\choose 2} = \frac{7\cdot 6}{2} = 21$ different ways to put five X's.

The probability that it does not rain on the last two days (I assume that is what you mean by 'weekend') is getting the first five X's and the last two empty. There is only one way to do that. Therefore, the probability that it would not rain is $\tfrac{1}{21}$ . Thus, the probability that it would rain is $\tfrac{20}{21}$ .

**mr fantastic** · January 31st 2009, 05:06 PM

Originally Posted by mathhomework

In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.

My take on this is the following:

Let X be the random variable number of days on the weekend that it rains.

X ~ Binomial(n = 2, p = 5/7).

Calculate Pr(X = 1).

**mr fantastic** · February 1st 2009, 12:11 AM

Originally Posted by mr fantastic

My take on this is the following:

Let X be the random variable number of days on the weekend that it rains.

X ~ Binomial(n = 2, p = 5/7).

Calculate Pr(X = 1).

My answer has assumed independence. If that's not the case, then more information is required to answer the question.

**Bilbo Baggins** · February 1st 2009, 08:47 AM

Originally Posted by ThePerfectHacker

Form a list of 7 empty spaces:
_ _ _ _ _ _ _

If it rains on Sunday, Monday, Tuesday, Wednesday, and Friday you write:
X X X X X _ _

There are ${7\choose 5} = {7\choose 2} = \frac{7\cdot 6}{2} = 21$ different ways to put five X's.

The probability that it does not rain on the last two days (I assume that is what you mean by 'weekend') is getting the first five X's and the last two empty. There is only one way to do that. Therefore, the probability that it would not rain is $\tfrac{1}{21}$ . Thus, the probability that it would rain is $\tfrac{20}{21}$ .

You are correct. However, that doesn't answer the question. The question is what is the probability that it will rain exactly ONE day on the weekend. You have determined the probability that it will rain on the weekend period. There are of course, three ways for that to happen: it can rain on Saturday, on Sunday, or on both. The number of total possible outcomes is simply 7!/5!/2! = 21 Fixing a rain day on Saturday and dry day on Sunday, we have 5!/4! = 5 outcomes. Doing the same for a rain day on Sunday and dry day on Sat. we have 5!/4!=5 outcomes. The probability is 10/21

**mr fantastic** · February 3rd 2009, 04:47 AM

Originally Posted by Bilbo Baggins

You are correct. However, that doesn't answer the question. The question is what is the probability that it will rain exactly ONE day on the weekend. You have determined the probability that it will rain on the weekend period. There are of course, three ways for that to happen: it can rain on Saturday, on Sunday, or on both. The number of total possible outcomes is simply 7!/5!/2! = 21 Fixing a rain day on Saturday and dry day on Sunday, we have 5!/4! = 5 outcomes. Doing the same for a rain day on Sunday and dry day on Sat. we have 5!/4!=5 outcomes. The probability is 10/21

I would dispute this answer. The probability of rain on any given day is 5/7. It's not a matter of listing outcomes .....

**Bruce** · February 3rd 2009, 09:31 AM

I think Bilbo is right. If it rains on saturday you have 4 pegs, let's say, to put on 5 different days, mon, tues, weds, thurs, friday.

there are 5 ways to do it as you can only miss one day out each time, and there are 5 days.

Then you double this as you could also put the first peg on sunday.

so you have the total number of ways you can get the outcome, divide it by the total number of outcomes and you have 10/21

**mr fantastic** · February 3rd 2009, 11:45 AM

Originally Posted by Bruce

I think Bilbo is right. If it rains on saturday you have 4 pegs, let's say, to put on 5 different days, mon, tues, weds, thurs, friday.

there are 5 ways to do it as you can only miss one day out each time, and there are 5 days.

Then you double this as you could also put the first peg on sunday.

so you have the total number of ways you can get the outcome, divide it by the total number of outcomes and you have 10/21

5 days out of 7 is only an average, it does not mean that it does rain 5 days a week. It only means that the probability of rain on a given day is 5/7. There's a big difference.

**Bilbo Baggins** · February 3rd 2009, 05:40 PM

Originally Posted by mr fantastic

5 days out of 7 is only an average, it does not mean that it does rain 5 days a week. It only means that the probability of rain on a given day is 5/7. There's a big difference.

True. However, I think that you are looking into the problem too deeply. A problem of this type usually doesn't require that type of scrutiny. I think that saying "rains an average of 5 times a week" is just code for "rains 5 times a week." Perhaps, because this is a statistics problem they didn't want to sound too unequivocal and added "on average"

**mathhomework** · February 3rd 2009, 11:29 PM

Thanks everybody for helping me!

**mr fantastic** · February 3rd 2009, 11:45 PM

Originally Posted by Bilbo Baggins

True. However, I think that you are looking into the problem too deeply. A problem of this type usually doesn't require that type of scrutiny. I think that saying "rains an average of 5 times a week" is just code for "rains 5 times a week." Perhaps, because this is a statistics problem they didn't want to sound too unequivocal and added "on average"

Then I suppose we will agree to disagree until the answer is given by the OP.

**Bruce** · February 5th 2009, 08:28 AM

My friend at school gave his take on the question:

there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49

the same for rain on sunday and none on saturday, so you end up with 20/49

this seems a plausible answer.

**Plato** · February 5th 2009, 09:53 AM

Originally Posted by Bruce

My friend at school gave his take on the question:
there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49 the same for rain on sunday and none on saturday, so you end up with 20/49. This seems a plausible answer.

I absolutely agree with that interpretation of this problem.
And I think that $\frac{10}{49}$ is correct.

**mr fantastic** · February 5th 2009, 03:29 PM

Originally Posted by Bruce

My friend at school gave his take on the question:

there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49

the same for rain on sunday and none on saturday, so you end up with 20/49

this seems a plausible answer.

Which exactly follows from what I posted way back in post #3 since $\Pr(X = 1) = ^2C_1 \left(\frac{5}{7}\right)^1 \left(\frac{2}{7}\right)^1 = \frac{20}{49}$ .

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