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In Rwanda, the chance for rain on any given day is 1/2. What

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In Rwanda, the chance for rain on any given day is 1/2. What [#permalink] New post 28 Feb 2005, 14:14
In Rwanda, the chance for rain on any given day is 1/2. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?

a) 4/7

b) 3/7

c) 35/128

d) 4/28

e) 28/135
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[#permalink] New post 28 Feb 2005, 14:20
The P will be C(7,4)* (1/2^7)

= 35/128

We have to select 4 rainy days from 7 total possible days and then multiply with the probability of each such selection.
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 [#permalink] New post 28 Feb 2005, 14:42
7C4 * (1/2)^7 = 35/128
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 [#permalink] New post 28 Feb 2005, 14:49
7C4/(2)^7 = 35/128
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 [#permalink] New post 28 Feb 2005, 16:03
Yes it is C

7C4 * 1/2^7
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 [#permalink] New post 28 Feb 2005, 16:34
How would you solve it if the percentage of rain likely on any days is 30% as against the given 50%( or 1/2).
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 [#permalink] New post 28 Feb 2005, 17:36
Yes, You guys are right.

To add a twist to the problem:

a) What is the probablility that it would rain on 4 consequitive days out of the 7 consequtive days?

b) What is the probablility that it would rain on only 3 alternate days out of the 7 consequtive days?

c) If it rains on the day before, the probability of it raining the next day becomes 1/4. What is the probability that it rains on 4 days out of 7 consecutive days in Rwanda?
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 [#permalink] New post 28 Feb 2005, 18:59
700Plus wrote:
How would you solve it if the percentage of rain likely on any days is 30% as against the given 50%( or 1/2).

Assuming your addition is a contineuation of the original question.
=7C4 * (3/10)^7 =(35x2187)/10^7
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 [#permalink] New post 28 Feb 2005, 19:14
nocilis wrote:
c) If it rains on the day before, the probability of it raining the next day becomes 1/4. What is the probability that it rains on 4 days out of 7 consecutive days in Rwanda?


=7C4 (1/2) (1/2*1/4) (1/2*1/4*1/4) (1/2*1/4*1/4*1/4)
=7C4 (1/2) (1/8) (1/32) (1/128)
=7C4 (1/2) (1/2)^3 (1/2)^5 (1/2)^7
=7C4 (1/2)^16

guys, pls do correct if any.
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 [#permalink] New post 28 Feb 2005, 19:22
nocilis wrote:
b) What is the probablility that it would rain on only 3 alternate days out of the 7 consequtive days?

=3*(1/2)^7 ??????????
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 [#permalink] New post 28 Feb 2005, 19:57
General rule

x successes out of n trials :

P(x)=n!/x!*(n-x)!*(p)^x*(1-p)^n-x

Some stuff from my Stats. class :wink:

P.S. BTW this my first post :wink:
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 [#permalink] New post 01 Mar 2005, 06:49
using Binomial. theorem
nCr (p)^r (q)^n-r

7C4 (1/2)^4 (1/2)^r

which is 35/128
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i'm stuck [#permalink] New post 01 Mar 2005, 09:35
can anyone please tell me why do we have to divide by seven?

My reasoning goest till we multiply for 1/2

I mean, it is a 7C4*1/2, but why divided by seven again?

thanks! :oops: :oops:
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 [#permalink] New post 01 Mar 2005, 10:30
OMG this is becoming such a good thought stimulating thread. I like the way you guys are thinking outside of the given question in order to test if you have really grasped the concept and mastered the approach. :good

Kevinw, it was not devided by seven. a^b means a to the power of b. In other words, for the original question people has multiplied C(7,4) by the probabilities of rain for four days and no rain for three days: (1/2)^4*(1/2)^3.
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 [#permalink] New post 01 Mar 2005, 14:07
MA wrote:
700Plus wrote:
How would you solve it if the percentage of rain likely on any days is 30% as against the given 50%( or 1/2).

Assuming your addition is a contineuation of the original question.
=7C4 * (3/10)^7 =(35x2187)/10^7


Using the following:

nPk = nCk * p^k * (1-p) ^ (n-k)
where nPk denotes the probability of an event having a given outcome exactly k times in n tests, p is the probability of the event having this outcome in each single test, and nCk comes from Combinations Theory.

it should be 7C4 * (3/10)^4 * (7/10)^3 iSN'T IT??????

CAN SOMEBODY TELL ME WHETHER I AM RIGHT OR WRONG?????
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 [#permalink] New post 01 Mar 2005, 14:46
Tajik4GMAT wrote:
General rule

x successes out of n trials :

P(x)=n!/x!*(n-x)!*(p)^x*(1-p)^n-x

Some stuff from my Stats. class :wink:

P.S. BTW this my first post :wink:


The way we solve it

7!/4!*(7-4)!*(1/2)^4*(1-1/2)^(7-4)

Am I right?
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 [#permalink] New post 01 Mar 2005, 15:06
Quote:
MA wrote:
Yes, You guys are right. To add a twist to the problem:
a) What is the probablility that it would rain on 4 consequitive days out of the 7 consequtive days?

=7P4(1/2^7)


MA, Hint: we are looking for 4 consequitive days -> There are only 4 sets of consequitve days Days {1,2,3,4} , {2,3,4,5}, {3,4,5,6}, {4,5,6,7}

Also, we don't care if it rains on the other 3 days or not.

Quote:
MA wrote:
b) What is the probablility that it would rain on only 3 alternate days out of the 7 consequtive days?

=3*(1/2)^7 ??????????

MA, Hint: 3 Alternate days are Days {1,3,5} or { 2,4,6} or {3,5,7}
Here, we don't want rain on other days.

Quote:
MA wrote:

c) If it rains on the day before, the probability of it raining the next day becomes 1/4. What is the probability that it rains on 4 days out of 7 consecutive days in Rwanda?


=7C4 (1/2) (1/2*1/4) (1/2*1/4*1/4) (1/2*1/4*1/4*1/4)
=7C4 (1/2) (1/8) (1/32) (1/128)
=7C4 (1/2) (1/2)^3 (1/2)^5 (1/2)^7
=7C4 (1/2)^16


Hint: We cannot use Binomial eqn here as the probabilities are dependant.
This one is definitely complicated - we may need to consider different cases such as consequitve days or alternate days and come up an answer.
I am sure somebody will solve this one.
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 [#permalink] New post 01 Mar 2005, 15:27
a) What is the probablility that it would rain on 4 consequitive days out of the 7 consequtive days?

4C1/(2)^7= 4/128 = 1/32

b) What is the probablility that it would rain on only 3 alternate days out of the 7 consequtive days?

3/128

c) If it rains on the day before, the probability of it raining the next day becomes 1/4. What is the probability that it rains on 4 days out of 7 consecutive days in Rwanda?

= 1 * 1/(2)^4 + 4 * (1/2 * 1/4)*(1/2^2) + 8 * (1/2 * 1/4 * 1/4) * 1/2 + 4 ( 1/2 * 1/4 * 1/4 * 1/4) = 11/32 ?.

there are 4 days of rain: 1 - it rains alternate days. 2 - it rains 2 consecutive days and 2 other days non consecutive 3 - it rains 3 days consec and 1 day non consec and 4 - it rains consec. Sum each of these probs to get the total prob. Not sure if this is the right approach, but thats my shot.
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 [#permalink] New post 01 Mar 2005, 15:34
just noticed nocilis' hints.

nocilis - for the 3 alternate q, isn't MA's response correct?.
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 [#permalink] New post 02 Mar 2005, 05:54
a) What is the probablility that it would rain on 4 consequitive days out of the 7 consequtive days?

4/7C4 ?

b) What is the probablility that it would rain on only 3 alternate days out of the 7 consequtive days?

3/2^5 ?
  [#permalink] 02 Mar 2005, 05:54
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