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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?

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?

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.

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?????

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?

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?

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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