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

How would you solve it if the percentage of rain likely on any days is 30% as against the given 50%( or 1/2).

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?

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

=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.

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

using Binomial. theorem
nCr (p)^r (q)^n-r

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

which is 35/128

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!

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.

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?????

The way we solve it

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

Am I right?

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.


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.



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.

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.

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 ?

Ok, let me try this out too.


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


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


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

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.

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

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