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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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using Binomial. theorem
nCr (p)^r (q)^n-r

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

which is 35/128
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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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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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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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 ?
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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
Ok, let me try this out too.

nocilis wrote:

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

Total outcome: 2^7
Possible outcomes for 4 consequitive days: 4 (1234, 2345, 3456, 4567)
Probability =4/2^7

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

Possible outcomes: 3 (135, 246, 357)
Probability=3/2^7

Quote:
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?

I don't know how to do this one. Anybody?
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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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Thanks for reminding about this thread.
I made these questions up, so there is no official answer.

My answer:
a) What is the probablility that it would rain on 4 consequitive days out of the 7 consequtive 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}

=4/ 2^7

b) What is the probablility that it would rain on only 3 alternate days out of the 7 consequtive days?
3 Alternate days are Days {1,3,5} or { 2,4,6} or {3,5,7}

= 3/2^7

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?

I think this one is tough, perhaps out of scope for GMAT - I will solve this one as time permits. As of now I don't have an answer.
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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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HongHu wrote:
Ok, let me try this out too.

nocilis wrote:

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

Total outcome: 2^7
Possible outcomes for 4 consequitive days: 4 (1234, 2345, 3456, 4567)
Probability =4/2^7

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

Possible outcomes: 3 (135, 246, 357)
Probability=3/2^7

Quote:
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?

I don't know how to do this one. Anybody?


q3---let's see, out of 4 possibilities 1234, 2345, 3456, 4567....

each scenario has p(e) = (1/2*1/4*1/4*1/4*1/2*1/2)

so p(e) = 4(1/2*1/4*1/4*1/4*1/2*1/2) = 1/128
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Re: In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
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nocilis wrote:
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


We can let R denote rain and N denote no rain; thus, we need to determine:

P(R-R-R-R-N-N-N) = (1/2)^7 = 1/128

However, there are 7C4 = 7!/(4! x 3!) = (7 x 6 x 5)/(3 x 2) = 35 ways to organize the letters R-R-R-R-N-N-N, so the total probability is 35 x 1/128 = 35/128.

Answer: C
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