Probability Problem - Cumbersome Solution

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Probability Problem - Cumbersome Solution

by student22 » Mon Dec 21, 2009 10:33 am

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If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5 day period?

A. 8/125

B. 2/25

C. 5/16

D. 8/25

E. 3/4
The solution manual states that the best way to solve this is to find the possible outcomes 2^5. Then you list the ways in which you can get rain on exactly 3 days as follows:

RRRNN; RRNRN; RRNNR...NNRRR will give you 10 desired outcomes. Then you divide 10/32 to get the answer.

Listing each desired outcome seems too time consuming and will easily eat into time better spent on more challenging problems. Does anybody have a faster way to solve this type of problem without listing each possible desired outcome?

Thanks!

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by Testluv » Mon Dec 21, 2009 12:49 pm
If you were dealing with only 2 or 3 days, then listing out the possibilities might be a good approach. But because we have 5 days, listing is definitely time-consuming and error-prone.

Here's a couple of other approaches.

For any question that involves a 50/50 probability between 2 events (rain/no rain, head/tail, etc), you can use the formula nCk/2^n where "n" is the number of times the event is being exercised and "k" represents desirable outcomes. "nCK" represents the number of ways the desirable event can occur while "2^n" is the total number of possibilities (so this is just the basic probability formula). The "2" comes from the fact that it is a binary event: rain/no rain. (So, if it were a die toss, we would use 6^n in the denominator.)

Here, we want rain on exactly 3 of the 5 days. So, n equals 5. We have: 5C3/2^5 = 5/16

You can also use Pascal's triangle on these problems:

.............................1 (n = 0)
..........................1.....1 (n = 1)
......................1......2........1 (n = 2)
..................1......3.......3.........1 (n = 3)
..............1......4.......6........4..........1 (n = 4)
..........1.......5.....10.....10.........5..........1 (n = 5)
.........(0)....(1)....(2)....(3).........(4).......(5) [5th row k values]
or
........(5).....(4)....(3)....(2).........(1).......(0) [5th row k values]
The triangle has many symmetrical properties that make it easy to remember. For example, each digit in a row (except for 1) is the sum of the digits directly above to the left and directly above to the right. Most GMAT questions will involve 3 or 5 tosses (or in this case, days). (So, you could just memorize the n =3 and n = 5 rows). Because there are 5 days, we go to the sixth row where n = 5. We want to know how it will rain on exactly three days. "10" is the third digit in the n = 5 row (whether you start counting from the left or the right). The sum of the digits in that row is 32. Therefore, the probability is 10/32 = 5/16.

EDIT: when counting, you need to start counting at zero. Looking at the bottom row, if we start from either the left or right, the "1" represents 1/32 chance of getting 0 rainy days (or ALL rainy days, which is just 0 non-rainy days); the "5" represents the chances of getting 1 rainy day out of 5 (or 1 non-rainy day or 1 head, etc). (So, the chances of getting exactly 1 rainy day are 5/32.) So, when I said "10" was the "third" digit, I should have made it clear that I meant zeroth, first, second, and third digit!

The position of each digit in a row you're looking at is just the k value you're interested in counting. After numerous attempts, I have edited the triangle to reflect this.
Last edited by Testluv on Mon Dec 21, 2009 11:59 pm, edited 4 times in total.
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by student22 » Mon Dec 21, 2009 3:17 pm
Testluv, thank you for your response, both methods are time savers.

Using the combination formula works very quickly and this was what I was looking for.

Also, I would have never thought to use pascal's triangle for this problem. It's a pretty good idea to memorize the 3rd and 5th row for problems like this.

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by Testluv » Mon Dec 21, 2009 3:25 pm
No problem.
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by valleeny » Mon Dec 21, 2009 4:34 pm
Essentially the logic for Testluv's formula is this.

Probability of rain or no rain on a given day =1/2
Probability of ANY combination of 3 days of rain out of 5 = (1/2)^3 * (1/2)^2 = 1/32
No. of ways to have 3 days rain out of 5 = 5C3 = 10
Total probabilty = 10 * 1/32

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by student22 » Mon Dec 21, 2009 6:29 pm
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?

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by Testluv » Mon Dec 21, 2009 7:07 pm
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.
Last edited by Testluv on Mon Dec 21, 2009 11:40 pm, edited 2 times in total.
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by Testluv » Mon Dec 21, 2009 9:39 pm
valleeny wrote:Essentially the logic for Testluv's formula is this.

Probability of rain or no rain on a given day =1/2
Probability of ANY combination of 3 days of rain out of 5 = (1/2)^3 * (1/2)^2 = 1/32
No. of ways to have 3 days rain out of 5 = 5C3 = 10
Total probabilty = 10 * 1/32
A very similar way of looking at the logic of the formula is this.

Probability = # desired outcomes/# of total possible outcomes

# desired outcomes = all the ways to get 3 days of rain out of 5 = 5C3 = 10

# of total possible outcomes = 2^5

Therefore, probability = 5C3/2^5.
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by Testluv » Tue Dec 22, 2009 12:07 am
For anyone who's further interested in the triangle approach, look here: https://www.beatthegmat.com/coin-flip-qu ... 17911.html

(I discovered that long before I joined this forum, Stuart explained the triangle in his typical awesome fashion!)
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by valleeny » Tue Dec 22, 2009 8:02 pm
Hi Testluv

The problem becomes more complicated if thr probability of raining is not equal to the probability of not raining = 1/2. In this case I don't think you can simply use (total desired events/total possible outcomes)?

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by Testluv » Tue Dec 22, 2009 8:22 pm
valleeny wrote:Hi Testluv

The problem becomes more complicated if thr probability of raining is not equal to the probability of not raining = 1/2. In this case I don't think you can simply use (total desired events/total possible outcomes)?
Yes, the math does get a bit tougher then. Supposing the probability of rain was 60%, and you wanted to know the likelihood of raining on exactly 3 (out of 5) days you would have: (.6)(.6)(.6)(.4)(.4) [ie, (.6)^3 * (.4)^2] but then you would have to consider the number of different arrangements which have 3 days of rain, so you would have to multiply that by 5C3 (or use the triangle at n value 5 and k value 3).

But, as you can see, the math here is very messy, and the GMAT is not about messy math.

I'm pretty sure this issue got discussed in the link I posted above because that thread goes on for pages.
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by valleeny » Tue Dec 22, 2009 8:34 pm
Hi Testluv

The problem becomes more complicated if thr probability of raining is not equal to the probability of not raining = 1/2. In this case I don't think you can simply use (total desired events/total possible outcomes)?

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by Testluv » Tue Dec 22, 2009 8:40 pm
valleeny wrote:Hi Testluv

The problem becomes more complicated if thr probability of raining is not equal to the probability of not raining = 1/2. In this case I don't think you can simply use (total desired events/total possible outcomes)?
?

Well, as I posted above, in that case it is best to approach the problem the way you first posted in this thread.

We're still sticking to the logic of the formula although you might say we're not using it directly; that is we're not contradicting the formula; we're just multiplying the number of ways the desired event can occur by the appropriate probability weighting. If you're asking me whether there is a way of doing it directly with the formula in one shot when the probability of rain and not rain is not equal, then, no, I don't know a way to do that.

When the probability is 1/2, no matter if we want 2, 3, 4 rainy days, we will always be raising 1/2 to 5, and so, approaching it the way you did in your first post, you will always come out with 1/32. (Of course, then we multiply by the appropriate number of ways, which we can get either from the nCk formula or the triagnle). For example, if we want rain exactly 2 days, then again we have (1/2)^2 * (1/2)^3, and because we are multiplying the same bases, we add the exponents, and we'll have (1/2)^5. But if we want rain on just 1 day, then we'll have (1/2)^1 * (1/2)^4, and again we'll have (1/2)^5. But if the probability of the desired event is anything other than 1/2, when we multiply (des) * (undes), we will always have different bases, and so the math is a lot more protracted.
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by viidyasagar » Fri Dec 25, 2009 2:12 am
Thanks a ton testluv..... i am just wondering whether all probability sums in this brilliant forum can be grouped (1 sum per concept) ..similarly for other chapters too.....ur explanations were lovely

Sorry if the above has already been done....somebody kindly guide a newbie....i am looking for topic wise posts in math....

Similarly concept wise posts in CR, i think these are the only 2 areas where topic wise grouping is possible.....

Tx

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by papgust » Fri Dec 25, 2009 3:09 am
This forum does not have topic wise posts in math. Infact, it isn't needed in this forum. You could use the search engine here. Say if you want posts regarding inequalities or triangles in data sufficiency, you could go to DS folder and do a search by providing the key word inequalities or triangles. In this way, you can look for the topic wise posts.