Problem Solving

If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?

(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

OA : later with explanation .

AIM GMAT wrote:If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?

(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

OA : later with explanation .

P = (good outcomes)/(total possible outcomes)

Let R = rain, N = no rain.

Total possible outcomes:
Each day could be R or N, giving us 2 possible outcomes for each day.
Over 5 days, total possible outcomes = 2*2*2*2*2 = 32.

Good outcomes:
Exactly 3 days of rain in a row:
RRRNN
RRRNR
NRRRN
NNRRR
RNRRR
5 good outcomes.

Exactly 4 days of rain in a row:
RRRRN
NRRRR
2 good outcomes.

Exactly 5 days of rain in a row:
RRRRR
1 good outcome.

Good outcomes = 5+2+1 = 8.

Thus, P(at least 3 days of rain in a row) = 8/32 = 1/4.

The correct answer is B.

GMATGuruNY wrote:

AIM GMAT wrote:If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?

(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

OA : later with explanation .

P = (good outcomes)/(total possible outcomes)

Let R = rain, N = no rain.

Total possible outcomes:
Each day could be R or N, giving us 2 possible outcomes for each day.
Over 5 days, total possible outcomes = 2*2*2*2*2 = 32.

Good outcomes:
Exactly 3 days of rain in a row:
RRRNN
RRRNR
NRRRN
NNRRR
RNRRR
5 good outcomes.

Exactly 4 days of rain in a row:
RRRRN
NRRRR
2 good outcomes.

Exactly 5 days of rain in a row:
RRRRR
1 good outcome.

Good outcomes = 5+2+1 = 8.

Thus, P(at least 3 days of rain in a row) = 8/32 = 1/4.

The correct answer is B.

I think you did a mistatke in counting the number of outcomes for 3 daus and 4 days. Actually, for exactly 3 days ----> there are 10 outcomes( 5!/(3!*2!)); for exactly 4 days, there are 5 outcomes ( 5!/(4!*1!)). The total is 10+5+1=16

Thus, P = 16/32 = 1/2. the answer is E.

Using pascal triangle amounts to the same result ( 1/2)

Is that correct??

My answer is 9/32. Please see below my way of reasoing. The key to note is raining at least 3 days "In a row"

1) At least 3 days in a row = In this case, we can consider three continous days raining as one unit and total distinct units wil be 3. We are arrange 3 units in 6 ways hence probability = 6/32.

2) At least 4 days in a row = In this case, we can consider four continous days raining as one unit and total distinct units wil be 2. We are arrange 2 units in 2 ways hence probability = 2/32.

3) At least 5 days in a row = In this case, we can consider five continous days raining as one unit and total distinct units wil be 1. We are arrange 1 unit in 1 way hence probability = 1/32.

Becuase all three cases above are mutually exclusive, the sum of probability rule will apply hence total probability of raning at least 3 days in a row will be 6/32 + 2/32 + 1/32 = 9/ 32 hence ANSWER = C

vaflaly wrote:

GMATGuruNY wrote:

AIM GMAT wrote:If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?

(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

OA : later with explanation .

P = (good outcomes)/(total possible outcomes)

Let R = rain, N = no rain.

Total possible outcomes:
Each day could be R or N, giving us 2 possible outcomes for each day.
Over 5 days, total possible outcomes = 2*2*2*2*2 = 32.

Good outcomes:
Exactly 3 days of rain in a row:
RRRNN
RRRNR
NRRRN
NNRRR
RNRRR
5 good outcomes.

Exactly 4 days of rain in a row:
RRRRN
NRRRR
2 good outcomes.

Exactly 5 days of rain in a row:
RRRRR
1 good outcome.

Good outcomes = 5+2+1 = 8.

Thus, P(at least 3 days of rain in a row) = 8/32 = 1/4.

The correct answer is B.

I think you did a mistatke in counting the number of outcomes for 3 daus and 4 days. Actually, for exactly 3 days ----> there are 10 outcomes( 5!/(3!*2!)); for exactly 4 days, there are 5 outcomes ( 5!/(4!*1!)). The total is 10+5+1=16

Thus, P = 16/32 = 1/2. the answer is E.

Using pascal triangle amounts to the same result ( 1/2)

Is that correct??

Your approach would be mathematically sound if the question asked for the probability that it will rain on at least 3 days.
But the problem above asks for the probability that it will rain at least 3 days IN A ROW.
If you revisit my solution above, all of the different ways are listed, so there is no room for error.

GMATGuruNY wrote:

vaflaly wrote:

GMATGuruNY wrote:

AIM GMAT wrote:If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?

(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

OA : later with explanation .

P = (good outcomes)/(total possible outcomes)

Let R = rain, N = no rain.

Total possible outcomes:
Each day could be R or N, giving us 2 possible outcomes for each day.
Over 5 days, total possible outcomes = 2*2*2*2*2 = 32.

Good outcomes:
Exactly 3 days of rain in a row:
RRRNN
RRRNR
NRRRN
NNRRR
RNRRR
5 good outcomes.

Exactly 4 days of rain in a row:
RRRRN
NRRRR
2 good outcomes.

Exactly 5 days of rain in a row:
RRRRR
1 good outcome.

Good outcomes = 5+2+1 = 8.

Thus, P(at least 3 days of rain in a row) = 8/32 = 1/4.

The correct answer is B.

I think you did a mistatke in counting the number of outcomes for 3 daus and 4 days. Actually, for exactly 3 days ----> there are 10 outcomes( 5!/(3!*2!)); for exactly 4 days, there are 5 outcomes ( 5!/(4!*1!)). The total is 10+5+1=16

Thus, P = 16/32 = 1/2. the answer is E.

Using pascal triangle amounts to the same result ( 1/2)

Is that correct??

Your approach would be mathematically sound if the question asked for the probability that it will rain on at least 3 days. Since the question asks for the probability that will rain on at least 3 days in a row, it's safer to write out the different ways in which a good outcome could be achieved. If you revisit my solution above, all the different ways are listed, so there is no room for error.

I understand my mystake now from your explaination. the key word is "in the row".
thank you

Hi Guru,

I see that you solved it by listing... is there another faster way to solve it using probability rules or combinations?

please advise

GMATGuruNY wrote:

vaflaly wrote:

GMATGuruNY wrote:

AIM GMAT wrote:If the probability of rain on any given day is 50%, what is the probability that it
will rain on at least 3 days in a row during a 5 day period?

(A) 3/32

(B) 1/4

(C) 9/32

(D)5/16

(E) 1/2

OA : later with explanation .

P = (good outcomes)/(total possible outcomes)

Let R = rain, N = no rain.

Total possible outcomes:
Each day could be R or N, giving us 2 possible outcomes for each day.
Over 5 days, total possible outcomes = 2*2*2*2*2 = 32.

Good outcomes:
Exactly 3 days of rain in a row:
RRRNN
RRRNR
NRRRN
NNRRR
RNRRR
5 good outcomes.

Exactly 4 days of rain in a row:
RRRRN
NRRRR
2 good outcomes.

Exactly 5 days of rain in a row:
RRRRR
1 good outcome.

Good outcomes = 5+2+1 = 8.

Thus, P(at least 3 days of rain in a row) = 8/32 = 1/4.

The correct answer is B.

I think you did a mistatke in counting the number of outcomes for 3 daus and 4 days. Actually, for exactly 3 days ----> there are 10 outcomes( 5!/(3!*2!)); for exactly 4 days, there are 5 outcomes ( 5!/(4!*1!)). The total is 10+5+1=16

Thus, P = 16/32 = 1/2. the answer is E.

Using pascal triangle amounts to the same result ( 1/2)

Is that correct??

Your approach would be mathematically sound if the question asked for the probability that it will rain on at least 3 days.
But the problem above asks for the probability that it will rain at least 3 days IN A ROW.
If you revisit my solution above, all of the different ways are listed, so there is no room for error.

Amrabdelnaby wrote:Hi Guru, I see that you solved it by listing... is there another faster way to solve it using probability rules or combinations?

please advise

For most test-takers, the solution above is likely to be the easiest and fastest approach.