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Director
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In Rwanda, the chance for rain on any given day is 1/2. What [#permalink]
 28 Feb 2005, 15:14
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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Manager
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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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SVP
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7C4 * (1/2)^7 = 35/128
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Manager
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7C4/(2)^7 = 35/128
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Director
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Yes it is C
7C4 * 1/2^7
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Manager
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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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Director
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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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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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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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SVP
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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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Manager
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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 
P.S. BTW this my first post
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Director
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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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Manager
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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! 
_________________
discipline is what I need.
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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. 
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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Manager
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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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Manager
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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 P.S. BTW this my first post The way we solve it
7!/4!*(7-4)!*(1/2)^4*(1-1/2)^(7-4)
Am I right?
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Director
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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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Manager
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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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Manager
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just noticed nocilis' hints.
nocilis - for the 3 alternate q, isn't MA's response correct?.
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Manager
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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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