Chances of raining = 50% therefore there are two options: either it rains or it does not rain.
Therefore the total options are 2^7 = 128 ways.
Now, say it rains on 4 days, so the number of ways it can rain on 4 given days = 7C4 = 7!/4! x 3! =35 ways.
Therefore probability is 35/128.
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winniethepooh
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knight247
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35/128 is correct. i have an issue though with 'the number of ways it can rain on 4 given days'
We have 4 rain days and 3 non rain days. In figures it would look as follows:
R R R R N N N Where 'R' stands for rain day and 'N' for non rain day
which can be arranged as
N N N R R R R
R R N N R R N
R N R N R N R
.
.
.
ETC
Shouldnt 'the number of ways it can rain on 4 given days' be 7P7 = 7!/(4!*3!) ???? Which btw is also equal to 35. Hope U can clarify this!!
We have 4 rain days and 3 non rain days. In figures it would look as follows:
R R R R N N N Where 'R' stands for rain day and 'N' for non rain day
which can be arranged as
N N N R R R R
R R N N R R N
R N R N R N R
.
.
.
ETC
Shouldnt 'the number of ways it can rain on 4 given days' be 7P7 = 7!/(4!*3!) ???? Which btw is also equal to 35. Hope U can clarify this!!
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winniethepooh
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Firstly, 7P7 is not equal to 35.
7P7 = 7!/(7-7)! = 7!/0! = 7!/1 = 7! ways.
Secondly, you should use permutation formula only when the order of the occurrence of an event is important.In this case however it does not matter when it rains, it can rain on any 4 days.
However, if the problem specified that it should rain on every alternate day then that would be a different scenario!
7P7 = 7!/(7-7)! = 7!/0! = 7!/1 = 7! ways.
Secondly, you should use permutation formula only when the order of the occurrence of an event is important.In this case however it does not matter when it rains, it can rain on any 4 days.
However, if the problem specified that it should rain on every alternate day then that would be a different scenario!
Last edited by winniethepooh on Mon Jun 20, 2011 3:01 am, edited 1 time in total.
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knight247
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7! is definitely not equal to 35. In case of R R R R N N N ....7P7 would be 7!/(4!*3!) because the Rs are repeated 4 times and the Ns are repeated 3 times. Which is equal to 35. Hope i clarified the point i'm trying to make.
And in this case isn't order important? R R R R N N N isn't the same as N N N R R R R
And in this case isn't order important? R R R R N N N isn't the same as N N N R R R R
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winniethepooh
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You're going on the wrong track buddy. Remember, always the formula for Permutations is: nPr = n!/r!x(n-r)!
For Combinations : nCr = n!/(n-r)!
Another point to remember is that a permutation of a certain number will always be higher than the combination of the same number. Hence, never can 7C7 be equal to 7P7!
Also,in this problem it doesn't matter if it rains in this pattern NNNNRRR versus in this pattern RRRRNNN.
For Combinations : nCr = n!/(n-r)!
Another point to remember is that a permutation of a certain number will always be higher than the combination of the same number. Hence, never can 7C7 be equal to 7P7!
Also,in this problem it doesn't matter if it rains in this pattern NNNNRRR versus in this pattern RRRRNNN.
Last edited by winniethepooh on Mon Jun 20, 2011 6:22 am, edited 1 time in total.
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winniethepooh
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Winnie - you have the right solution but the question does not say 4 consecutive days. It says any 4 days out of 7 consecutive days. That is the reason we take 7C4. If it had been 4 consecutive days, the approach would be different.
Another way of looking at it would be as that of rearranging the letters in a word to form a new word. Suppose the word is RRRRNNN. There are a total of 7 letters of which 4 are identical and 3 are identical. Therefore the number of ways to rearrange will be 7!/(4!*3!) which is same as 7C4.
Hope this helps.
Another way of looking at it would be as that of rearranging the letters in a word to form a new word. Suppose the word is RRRRNNN. There are a total of 7 letters of which 4 are identical and 3 are identical. Therefore the number of ways to rearrange will be 7!/(4!*3!) which is same as 7C4.
Hope this helps.
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winniethepooh
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