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# Full Length Practice Test: Math # 26: pg 409: I understand

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Manager
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Full Length Practice Test: Math # 26: pg 409: I understand [#permalink]

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29 Aug 2006, 07:50
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Full Length Practice Test: Math
# 26: pg 409: I understand the solution given in the back. My question is:
They enlist desired outcomes one by one, which is slow and tedious. Is there a smarter/better way to do thisâ€¦?
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. Â¾
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Saumil Annegiri,

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Re: Kaplan Premier -> Question (Math) [#permalink]

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29 Aug 2006, 10:17
saumster wrote:
Full Length Practice Test: Math
# 26: pg 409: I understand the solution given in the back. My question is:
They enlist desired outcomes one by one, which is slow and tedious. Is there a smarter/better way to do thisâ€¦?
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. Â¾

Yup there is. It's called the binomial theorem. And it works perfectly for these types of questions. It works under the condition that there are only two outcomes such as in this case whether it rains or not. Each trial has to be independent of the previous trial. In other words you can say with replacement.

The formula is C(n,r) P(x)^r (1-P(x))^n-r

In this case 5C3 x (1/2)^3 (1/2)^2= 5/16

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29 Aug 2006, 10:28
Personally I've found it helps to already be familiar with the scenarios involving heads and tails for instance, up to 4 coins - this can be used in pretty much any 50% chance scenario.

How many have 2 heads (and 2 tails) only (6), how many have 3 heads (4) or 3 tails (4) and 4 heads (1) and 4 tails (1). Making a grand total of 16.

But in any case even using my scenario, I get 12/32, which gives me 3/8, not in the answers!

Where did I go wrong?
MG

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29 Aug 2006, 10:39
In dealing with probabiities it helps to just think in terms of possible number of total outcomes

So for this case its 2^5 since each day has two possible outcomes and there are 5 days.

For the numerator you can manually count the number of outcomes given the condition in the question stem.

In this case it can be ( R for Rain NR no rain)

R,R,R,NR,NR
R,R,NR, R,R ....and so forth an easier way of arriving at the total for the numerator is to use combinations so in this case you are counting the number of ways it can rain 3 times out of the five days so that is 5 combination 3 so answer 5C3/2^5

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29 Aug 2006, 11:41
apollo168 wrote:
In dealing with probabiities it helps to just think in terms of possible number of total outcomes

So for this case its 2^5 since each day has two possible outcomes and there are 5 days.

For the numerator you can manually count the number of outcomes given the condition in the question stem.

In this case it can be ( R for Rain NR no rain)

R,R,R,NR,NR
R,R,NR, R,R ....and so forth an easier way of arriving at the total for the numerator is to use combinations so in this case you are counting the number of ways it can rain 3 times out of the five days so that is 5 combination 3 so answer 5C3/2^5

Great explanation! When do you use combinations to arrive at the numerator though? I have found sometimes it doesnt work. Im a novice at Probability types.

Also I didn't follow the Binomial part you explained above...?
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29 Aug 2006, 21:59
Another way I sometimes look at this is that we know the total number of outcomes is 64. Therefore you can rule out all but 5/16 because none of the other answers match up.

This doesn't always work, but does anyone else ever do this?

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30 Aug 2006, 00:06
5C3 * (1/2)^3 * (1/2)^2 = 5/16

Use binomial theorem.
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30 Aug 2006, 03:23
agree with psd.

use binomial theorem to get 5!(3!2!) X 0.5^3. 0.5^2 = 10/32 = 5/16

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Re: Kaplan Premier -> Question (Math) [#permalink]

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30 Aug 2006, 06:18
apollo168 wrote:
saumster wrote:
Full Length Practice Test: Math
# 26: pg 409: I understand the solution given in the back. My question is:
They enlist desired outcomes one by one, which is slow and tedious. Is there a smarter/better way to do thisâ€¦?
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. Â¾

Yup there is. It's called the binomial theorem. And it works perfectly for these types of questions. It works under the condition that there are only two outcomes such as in this case whether it rains or not. Each trial has to be independent of the previous trial. In other words you can say with replacement.

The formula is C(n,r) P(x)^r (1-P(x))^n-r

In this case 5C3 x (1/2)^3 (1/2)^2= 5/16

Definately a time saving theory. Just curious if there is another similiar shortcut with dependent probability, and if the theorum works with at least/most questions?

I solved this old fashioned style:

5C3/(1/2)^5 ---> 5/16

What if the question asked, what is the probability that a coin comes up heads exactly 6 out of 10 tosses?

---> 10C6*(1/2)^6*(1/2)^4 ---> 55/128 ???

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30 Aug 2006, 06:25
Yup the theorem works for at least at most problems as well. In the same way we would subtract the alternative probability from 1 the old fashion way. So in this case if the question were to ask if the probability that it were to rain 1 out of 5 days,

1- P(0)= 1- 5C0 (1/2)^0 (1/2)^5

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30 Aug 2006, 06:36
apollo168 wrote:
Yup the theorem works for at least at most problems as well. In the same way we would subtract the alternative probability from 1 the old fashion way. So in this case if the question were to ask if the probability that it were to rain at least 1 out of 5 days,

1- P(0)= 1- 5C0 (1/2)^0 (1/2)^5

You meant at least, right?

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30 Aug 2006, 06:41
GMATT73 wrote:
apollo168 wrote:
Yup the theorem works for at least at most problems as well. In the same way we would subtract the alternative probability from 1 the old fashion way. So in this case if the question were to ask if the probability that it were to rain at least 1 out of 5 days,

1- P(0)= 1- 5C0 (1/2)^0 (1/2)^5

You meant at least, right?

Yup at least sorry. Didnt see this one

What if the question asked, what is the probability that a coin comes up heads exactly 6 out of 10 tosses?

---> 10C6*(1/2)^6*(1/2)^4 ---> 55/128 ???
yup this is correct. And the binomial theorem applies to problems with unequal probabilities as well if the coin or dice were to be loaded. Say the probability of it turning up heads is 2/3 then 10C6*(2/3)^6*(1/3)^4

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15 Sep 2006, 11:21
Can you please suggest resources / links (I have done most of the book salready) for "Binomial Theorem"? I am not familiar with it.

apollo168 wrote:
GMATT73 wrote:
apollo168 wrote:
Yup the theorem works for at least at most problems as well. In the same way we would subtract the alternative probability from 1 the old fashion way. So in this case if the question were to ask if the probability that it were to rain at least 1 out of 5 days,

1- P(0)= 1- 5C0 (1/2)^0 (1/2)^5

You meant at least, right?

Yup at least sorry. Didnt see this one

What if the question asked, what is the probability that a coin comes up heads exactly 6 out of 10 tosses?

---> 10C6*(1/2)^6*(1/2)^4 ---> 55/128 ???
yup this is correct. And the binomial theorem applies to problems with unequal probabilities as well if the coin or dice were to be loaded. Say the probability of it turning up heads is 2/3 then 10C6*(2/3)^6*(1/3)^4

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15 Sep 2006, 12:16
Success = Rains 3 days and doesn't rain for the other 2.

P(Success) = (1/2 x 1/2 x 1/2) x (1/2x1/2)
= 1/32

Number of ways of picking 3 days of 5 = 5x4x3/3x2 = 10

Total probabibility = 10/32 = 5/16

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15 Sep 2006, 12:16
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# Full Length Practice Test: Math # 26: pg 409: I understand

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