Problem Solving

In Rwanda, the chance for rain on any given day is 50%. 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

Good one....

Ans C
Basically you have to arrange 4 Y(to rain) and 3N (not to rain)
= 7!/(4!*3!) = 35

Hence prob = 35/2^7 = 35/128

arora007 wrote:In Rwanda, the chance for rain on any given day is 50%. 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

Good one....

4 days can be picked from 7 days in 7c4 ways = 35 possible ways.
On a single selection, probability of raining is 1/2 * 1/2 * 1/2 *1/2 * 1/2 * 1/2 * 1/2 = 1/(2^7)
So total = 35/128

arora007 wrote:In Rwanda, the chance for rain on any given day is 50%. 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

Good one....

The "consecutive" bit threw me off. I assumed that they meant that it rains on 4 consecutive days. In that case, I found the different ways it could rain for 4 consecutive days, which was 24 (I calculated different arrangements of 7, treating the four days as one). But it actually means any 4 days out of 7 consecutive days. So in that cays, we pick 4 out of 7 and get 35.

kvcpk wrote:

arora007 wrote:In Rwanda, the chance for rain on any given day is 50%. 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

Good one....

4 days can be picked from 7 days in 7c4 ways = 35 possible ways.
On a single selection, probability of raining is 1/2 * 1/2 * 1/2 *1/2 * 1/2 * 1/2 * 1/2 = 1/(2^7)
So total = 35/128

I don't completelty:
7! / 4! * 3!
7 is total amount of options
4 is how many options u need in a group
3 is the remaining options
you get 35 total combinations that it will rain 4 days in any 7 of those days
then u need total chance
50% per day over 7 days
0.5 ^ 7
1/128
35 combinations * 1/128 probability it will rain all 7 days
.

is this the right logic?

phoenixhazard wrote:

kvcpk wrote:

arora007 wrote:In Rwanda, the chance for rain on any given day is 50%. 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

Good one....

4 days can be picked from 7 days in 7c4 ways = 35 possible ways.
On a single selection, probability of raining is 1/2 * 1/2 * 1/2 *1/2 * 1/2 * 1/2 * 1/2 = 1/(2^7)
So total = 35/128

I don't completelty:
7! / 4! * 3!
7 is total amount of options
4 is how many options u need in a group
3 is the remaining options
you get 35 total combinations that it will rain 4 days in any 7 of those days
then u need total chance
50% per day over 7 days
0.5 ^ 7
1/128
35 combinations * 1/128 probability it will rain all 7 days
.

is this the right logic?

Let R = rain, N = no rain.

P(RRRRNNN) = (1/2)^7 = 1/128.

We need to multiply this result by the number of ways to arrange RRRRNNN in order to account for ALL the different ways we could get 4 days of R.

The number of ways to arrange 7 elements is 7!. But when we have repeated elements such as RRRR and NNN, the number of unique arrangements will be reduced. To account for the four R's, we need to divide by 4!, and to account for the 3 N's, we need to divide by 3!: 7!/(4!*3!) = 35.

Thus, P(4 days of R) = 35*(1/128) = 35/128.

Does this help?

GMATGuruNY wrote:

phoenixhazard wrote:

kvcpk wrote:

arora007 wrote:In Rwanda, the chance for rain on any given day is 50%. 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

Good one....

4 days can be picked from 7 days in 7c4 ways = 35 possible ways.
On a single selection, probability of raining is 1/2 * 1/2 * 1/2 *1/2 * 1/2 * 1/2 * 1/2 = 1/(2^7)
So total = 35/128

I don't completelty:
7! / 4! * 3!
7 is total amount of options
4 is how many options u need in a group
3 is the remaining options
you get 35 total combinations that it will rain 4 days in any 7 of those days
then u need total chance
50% per day over 7 days
0.5 ^ 7
1/128
35 combinations * 1/128 probability it will rain all 7 days
.

is this the right logic?

Let R = rain, N = no rain.

P(RRRRNNN) = (1/2)^7 = 1/128.

We need to multiply this result by the number of ways to arrange RRRRNNN in order to account for ALL the different ways we could get 4 days of R.

The number of ways to arrange 7 elements is 7!. But when we have repeated elements such as RRRR and NNN, the number of unique arrangements will be reduced. To account for the four R's, we need to divide by 4!, and to account for the 3 N's, we need to divide by 3!: 7!/(4!*3!) = 35.

Thus, P(4 days of R) = 35*(1/128) = 35/128.

Does this help?

Yep thanks. I had to think about it in a different way:
35 is the total number of possible combinations that it can rain for 4 days out of 7. Since we are using every single one of those 7 days to make the 35 combinations then we must see the probility that it can rain on every single one of those days which is 1/128. Multiply them together to see the probility that one of those combinations will occur with the given probability.

akdayal wrote:Ans C
Basically you have to arrange 4 Y(to rain) and 3N (not to rain)
= 7!/(4!*3!) = 35

Hence prob = 35/2^7 = 35/128

thank u........