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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Probability of rain
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wilderness
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Ian Stewart
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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.
(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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muzali
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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
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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muzali
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Sorry, had not seen the 20% question...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.
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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jimmiejaz
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Muzali,muzali wrote:Sorry, had not seen the 20% question...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.
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
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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tohellandback
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I have seen this answer in different threads but still don't understand why do you calculate the arrangement and then multiply.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
could you please explain?
The powers of two are bloody impolite!!
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Robinmrtha
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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
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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goalevan
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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%
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%
