Probability Question

LatecomerX SG

Given that the chances of raining on a day is 90%, what is the probability of having 4 rainy days within a period of one week?

wee_ws SG

Hi,

Consider the use of the Binomial distribution. Thanks.

Cheers,
Wen Shih

LatecomerX SG

Hi,

Consider the use of the Binomial distribution. Thanks.

Cheers,
Wen Shih

Thanks for the reply, Wen Shih.

I looked up the term in Wikipedia and found this PMF:

http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function

It works correctly when the chance of raining is 50% (original question), which gives the answer of 7C4 / 2^4 * 2^3 = 35/128, but when it becomes 90%, the probability is lesser than the original answer (7C4 * 9^4 / 10^4 * 10^3 ~= 0.023), which does not make sense. Any idea where I did a mistake when applying the formula? Or did I use the wrong formula after all?

Fortmax777 SG

(7C4 * 9^4 / 10^4 * 10^3 ~= 0.023),

Think you made a mistake:

In your case n = 7, k = 4 , p = 0.9 (sucess is a raining day)

Therefore use: 7C4*0.9^4*0.1^3.

Cannot evaluate as I do not have a caluclator with me but I am sure this will give you a reasonable answer.

LatecomerX SG

(7C4 * 9^4 / 10^4 * 10^3 ~= 0.023),

Think you made a mistake:

In your case n = 7, k = 4 , p = 0.9 (sucess is a raining day)

Therefore use: 7C4*0.9^4*0.1^3.

Cannot evaluate as I do not have a caluclator with me but I am sure this will give you a reasonable answer.

Thanks for your input.

I missed out a pair of parenthesis in the working above (it should be (7C4 * 9^4 / [10^4 * 10^3] ~= 0.023)), but my final answer is based on the one I had on paper that is correct.

7C4*0.9^4*0.1^3

= 7C4 * (9/10)^4 * (1/10)^3

= 7 * 6 * 5 / 1 / 2 / 3 * 9^4 / 10^4 / 10^3

= 7 * 5 * 6 / 6 * 9^4 / 10^7

= 35 * 9^4 / 10^7

~= 0.023

As you can see, just by dividing by 10^7 alone would make the probability very small.

Whereas when the chance of rain is 50%, it's 35/128

~= 0.273

TrueHeart SG

Thanks for your input.

I missed out a pair of parenthesis in the working above (it should be (7C4 * 9^4 / [10^4 * 10^3] ~= 0.023)), but my final answer is based on the one I had on paper that is correct.

7C4*0.9^4*0.1^3

= 7C4 * (9/10)^4 * (1/10)^3

= 7 * 6 * 5 / 1 / 2 / 3 * 9^4 / 10^4 / 10^3

= 7 * 5 * 6 / 6 * 9^4 / 10^7

= 35 * 9^4 / 10^7

~= 0.023

As you can see, just by dividing by 10^7 alone would make the probability very small.

Whereas when the chance of rain is 50%, it's 35/128

~= 0.273

Hi LatecomerX,

I am not very good at Maths, so please do forgive me if my suggestion is incorrect. Too much focus have been paid on the four days of rain, to the point that you forget that there are 3 days of no rain. There is only a 10% chance that there will be no rain for a day, and for there to be no rain for 3 days, the odds will have to be really low.

This is in contrast to there being 50% chance of rain. This means very directly ( and literally ) that there will either be rain or no rain, thus the odds are much higher, for there to be 4 days with rain and 3 days without rain.

Cheers.

Fortmax777 SG

Hi LatecomerX,

Cheers.

Hi TrueHEart is right.

The Binomial Distribution gives the probability of there being exactly 4 days of rain and 3 days of no rain.

If you want the probability of at least four days of rain and letting X be the RV that reps no of rainy days then P(X <= 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4). Then P(X>4) = 1 - P(X<=4). i.e you have to apply the binomial distribution 4 times.

LatecomerX SG

Hi LatecomerX,

Cheers.

I did miss out the fact that it's much harder to get 3 days without rain when the chance of rain is 90%. Thanks for the excellent explanation.