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Probability of rain

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Probability of rain

by wilderness » Fri Jul 25, 2008 9:56 am
A question with answer is given below.

How can we solve if the probability of rain on a given day is not 50%. Let say the queston was

"If the probability of rain on a certain day is 20%, what is the probability that it will rain in exactly 3 out of a 5-day period ? "

Will we only divide our current results by 2.5 since 50/20 is 2.5 or is there some other method.
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by Ian Stewart » Fri Jul 25, 2008 10:31 am
That makes the problem a bit more difficult. Say, first, you wanted to know the probability of rain on the first three days, then no rain on the last two days, and the probability of rain is 0.2. Then the probability would be:

(0.2)^3 * (0.8)^2

Of course, the probability will be the same for any other particular sequence of three rainy days and two non-rainy days. As in the problem above, there are 5C3 = 10 different sequences, so we need to multiply the above by 10.

10* (0.2)^3 * (0.8)^2 should be the answer, if you change the probability of rain to 0.2. Mind you, I'd be a bit surprised to see that on a GMAT- it's not completely out of the question, but it's unlikely.
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I am still confused

by saabidi » Wed Nov 26, 2008 1:05 pm
Hi Guys,

Can you please explain how # of desired outcome is being calculated.
I am having hard time understanding how the answer is 10?

Thanks
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I am still confused

by saabidi » Wed Nov 26, 2008 1:07 pm
Hi Guys,

Can you please explain how # of desired outcome is being calculated.
I am having hard time understanding how the answer is 10?

Thanks
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by muzali » Wed Nov 26, 2008 2:58 pm
Probability of rain on any day=50% = 1/2
Probability of rain on 3 days out of 5 i.e. it rains on 3 days and does not rain on 2 days= [(1/2)^3]*[(1/2)^2] = (1/2)^5=1/32

Number of ways you can arrange these three days on which it will rain out of a total of five days = 5C3 = 5*4/2 = 10

Therefore, the required probability is = (1/32)*10 = 10/32=5/16
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Re: Probability of rain

by muzali » Sat Nov 29, 2008 5:03 pm
wilderness wrote:A question with answer is given below.

How can we solve if the probability of rain on a given day is not 50%. Let say the queston was

"If the probability of rain on a certain day is 20%, what is the probability that it will rain in exactly 3 out of a 5-day period ? "

Will we only divide our current results by 2.5 since 50/20 is 2.5 or is there some other method.
Sorry, had not seen the 20% question...
the appraoch is the same as 50%

Probability of rain on any day=20% = 1/5
Probability of rain on 3 days out of 5 i.e. it rains on 3 days and does not rain on 2 days= [(1/5)^3]*[(1/5)^2] = (1/5)^5

Number of ways you can arrange these three days on which it will rain out of a total of five days = 5C3 = 5*4/2 = 10

Therefore, the required probability is = ((1/5)^5)*10 = 2/5^4 =2/625
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Re: Probability of rain

by jimmiejaz » Mon Dec 01, 2008 6:22 am
muzali wrote:
wilderness wrote:A question with answer is given below.

How can we solve if the probability of rain on a given day is not 50%. Let say the queston was

"If the probability of rain on a certain day is 20%, what is the probability that it will rain in exactly 3 out of a 5-day period ? "

Will we only divide our current results by 2.5 since 50/20 is 2.5 or is there some other method.
Sorry, had not seen the 20% question...
the appraoch is the same as 50%

Probability of rain on any day=20% = 1/5
Probability of rain on 3 days out of 5 i.e. it rains on 3 days and does not rain on 2 days= [(1/5)^3]*[(1/5)^2] = (1/5)^5

Number of ways you can arrange these three days on which it will rain out of a total of five days = 5C3 = 5*4/2 = 10

Therefore, the required probability is = ((1/5)^5)*10 = 2/5^4 =2/625
Muzali,
Just to correct you....
The probability that it does not rain on any day is 1-1/5 = 4/5
so, as Ian explained above probability will be
[(1/5)^3]*[(4/5)^2]*10
hope it helps...
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by tohellandback » Tue Jun 30, 2009 11:07 pm
muzali wrote:Probability of rain on any day=50% = 1/2
Probability of rain on 3 days out of 5 i.e. it rains on 3 days and does not rain on 2 days= [(1/2)^3]*[(1/2)^2] = (1/2)^5=1/32

Number of ways you can arrange these three days on which it will rain out of a total of five days = 5C3 = 5*4/2 = 10

Therefore, the required probability is = (1/32)*10 = 10/32=5/16
I have seen this answer in different threads but still don't understand why do you calculate the arrangement and then multiply.
could you please explain?
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by Robinmrtha » Tue Jun 30, 2009 11:30 pm
the probability of raining is 1/2
so the probability that it will rain exactly three days in five days will be
rains in 1st day x rains in 2nd day x rains in 3rd day x doesnt rain in 4th day x doesnt rain in 5th day + ..........(all possible arrangements of three)
i.e. total ways 3 days can be selected from 5
or 5C3=10
now since the probability is half...
the probability for raining and not raining both is same i.e. 1/2
so
so for any combination of days the probability will be same
i.e 1/2x1/2x1/2x1/2x1/2=1/32
And probability will repeat for all possible combination. i.e. 10
so multiply by 10
i.e. 10x 1/32 =5/16
hope this helps
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by goalevan » Mon Mar 01, 2010 6:27 pm
The binomial distribution formula comes in handy for this problem. Basically what most of the previous posts have contained:

p Probability of success (rain): 1/5
n Number of trials: 5
k Number of successes: 3

Binomial distribution formula: nCk * p^k * (1-p)^(n-k)

In other words, the possible combinations of k successes in n trials (different ways to order those k items) multiplied by the probability of one of those strings occurring.

nCk = 10
p^k = (1/5)^3 = 1/125
(1-p)^(n-k) = (4/5)^2 = 16/25
10 * 1/125 * 16/25 = 32/625 = 5.12%
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