IMO B

Total Possible Outcomes -> 2^5 = 32

Now types of combination when it rains on at least 3days in a row

RRRRR
NRRRR
RRRRN
RNRRR
RRRNR
RRRNN
NRRRN
NNRRR

Total = 8

therefore probability of 3 continuous days of rain = 8/32 = 1/4

You probably got E because you were solving for "what's the probability that it rains at least 3 days out of the five?" In a row changes the questions.

Listing out the cases as Sourabh did is probably the easiest and quickest way to solve it. You could also use a probability tree, but that'll get messy with 5 iterations.

SoCan wrote:You probably got E because you were solving for "what's the probability that it rains at least 3 days out of the five?" In a row changes the questions.

Listing out the cases as Sourabh did is probably the easiest and quickest way to solve it. You could also use a probability tree, but that'll get messy with 5 iterations.

Yes, i was trying to solve the question for "what's the probability that it rains at least 3 days out of the five?"

But thanks to both of you for replying.

Regards
Vinni

the easiest approach in at least probablility is

1-non favourable outcomes :

what is non favourable ?
1>it rains exactly 1 day - > P=(1/2)
2>it rains exactly 2 days- > P=(1/2*1/2)

add both we get 1/2+1/4 = 3/4

hence the answer follows :
1-3/4 = 1/4

cheers !!

try this now people :


The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors ?

oa-[spoiler](3*.63)[/spoiler]

sorry for the wrong answer-, its [spoiler].063*3[/spoiler]

Hi bblast

I have a small issue with your approach.

The question asks for the probability that it will rain on at least 3 days in a row during a 5 day period

Also, IMO
For Non Favorable outcomes that it will rain on any three days will not be 3/4

Probability that it rains exactly on no days = 1/32
Probability that it rains exactly one day = 5/32
Probability that it rains exactly 2 days will be = 10/32

It will be 1/32+5/32+10/32 = 16/32 = 1/2

therefore probability that it will rain on any three day = 1-1/2 = 1/2


Now, regarding your question are you sure that the answers is 3x0.69 and not 0.189 (0.3 x 0.63)

IMO

Probability that a visitor will buy a candy -> 0.3
Therefore the probability that a visitor will not buy a candy -> 1-0.3 = 0.7

Now we have 3 visitors, out of which two buys the candy. Also the arrangement matters for this problem

so possible solutions will be

(0.3 X 0.3 x 0.7) + (0.3 x 0.7 x 0.3) + (0.7 X 0.3 X 0.3)

= 3x7X3x3/1000
= 189/1000
= 0.189

good work, the answer is indeed what u found out, recheck whats in my spoiler, seems u missed it.

bblast wrote:try this now people :


The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors ?

oa-[spoiler](3*.63)[/spoiler]

your spoiler says 3*.63 = 1.89 and not 0.189, which is the answer.

bblast wrote:the easiest approach in at least probablility is

1-non favourable outcomes :

what is non favourable ?
1>it rains exactly 1 day - > P=(1/2)
2>it rains exactly 2 days- > P=(1/2*1/2)

add both we get 1/2+1/4 = 3/4

hence the answer follows :
1-3/4 = 1/4

cheers !!

While the answer is correct, the methodology and even the calculations within that methodology are incorrect. On similar questions, you may not be as lucky in stumbling on the right answer.

When looking at sets of events and calculating probabilities, make sure to take each event in the set under consideration.

bblast wrote:the easiest approach in at least probablility is

1-non favourable outcomes :

what is non favourable ?
1>it rains exactly 1 day - > P=(1/2)
2>it rains exactly 2 days- > P=(1/2*1/2)

add both we get 1/2+1/4 = 3/4

hence the answer follows :
1-3/4 = 1/4

cheers !!

Be careful, bblast. You were lucky.
For many problems, the following strategy works well:

P(good outcomes) = 1 - P(bad outcomes).

But your reasoning overlooks quite a few bad outcomes, bolded in red below:

P(at least 3 days of rain in a ROW) = 1 - P(no rain) - P(exactly 1 day of rain) - P(exactly 2 days of rain) - P(exactly 3 days of rain NOT in a row) - P(exactly 4 days of rain without 3 days of rain in a row)

Your calculations are not quite correct. If R=rain and N=no rain:
P(exactly 1 R):
P(RNNNN) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.
Since R can occur on any of the 5 days, we must multiply the result above by 5:
P(exactly 1 R) = 5*(1/32) = 5/32.

P(exactly 2 R):
P(RRNNN) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.
The number of ways to arrange RRNNN = 5!/(2!3!) = 10. Thus, to account for all the ways in which we could get two R's, the result above must be multiplied by 10.
P(exactly 2 days of rain) = 10*(1/32) = 10/32.

Subtracting these fractions from 1, we get:
1 - 5/32 - 10/32 = 17/32, which is not the correct answer.

So please be careful. Your approach will work well for many problems, but it's important to count ALL the different ways a bad outcome could occur.

In this particular problem, I think it's easier to determine P(at least 3 R in a row) directly:

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.

bblast wrote:good work, the answer is indeed what u found out, recheck whats in my spoiler, seems u missed it.

Your spoiler says that the chance of this happening is greater than 100%, which is impossible.
The answer is indeed 3(0.3*0.3*0.7)=3(0.063)=0.189.

SoCan wrote:

bblast wrote:good work, the answer is indeed what u found out, recheck whats in my spoiler, seems u missed it.

Your spoiler says that the chance of this happening is greater than 100%, which is impossible.
The answer is indeed 3(0.3*0.3*0.7)=3(0.063)=0.189.

looks like m taking quite a bit of stick on this one. come on guys- lemme go now. I erred.

Thanks for polishing this up Mitch.

sourabh33 wrote:Hi bblast

I have a small issue with your approach.

The question asks for the probability that it will rain on at least 3 days in a row during a 5 day period

Also, IMO
For Non Favorable outcomes that it will rain on any three days will not be 3/4

Probability that it rains exactly on no days = 1/32
Probability that it rains exactly one day = 5/32
Probability that it rains exactly 2 days will be = 10/32

It will be 1/32+5/32+10/32 = 16/32 = 1/2

therefore probability that it will rain on any three day = 1-1/2 = 1/2

I was exactly doing the same thing, ignoring in a row phrase, which got me 1/2 as an answer.

Thank you very much guys for this fruitful discussion.

Regards
Vinni