If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
Please explain, thank
OA : [spoiler]5/16[/spoiler]
Source : Manhattan challenge problem
chicago rainy days
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The problem tells you that you have exactly 1/2 probability of an event happening, which is basically a coin-toss probability.
Now, in order to get rain three days out of five in a specific arrangement (i.e. first three days rain, next two days sunshine or first two days sunshine and next three days rain), the probability will be (1/2)^5 = 1/32. But, since we're not interested in the exact order of rain vs. sunshine days, all possible arrangements are good. This is why you need to determine all possible combinations of this event.
Examples:
R R S S S
R S S R R
R S S S R
and so on and so forth.
The exact number of possible cases will be 5C3, since you've got five possible slots or days for your three rainy days. 5C3 = 10, so the probability you're looking for is:
10/32 = 5/16.
Now, in order to get rain three days out of five in a specific arrangement (i.e. first three days rain, next two days sunshine or first two days sunshine and next three days rain), the probability will be (1/2)^5 = 1/32. But, since we're not interested in the exact order of rain vs. sunshine days, all possible arrangements are good. This is why you need to determine all possible combinations of this event.
Examples:
R R S S S
R S S R R
R S S S R
and so on and so forth.
The exact number of possible cases will be 5C3, since you've got five possible slots or days for your three rainy days. 5C3 = 10, so the probability you're looking for is:
10/32 = 5/16.
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So from July 4th to July 8th means.. 5 days.
No of ways of choosing three days out of five is 5C3 = 10
probability of rain on a given day = 1/2
so probability of rain not happening on a given day is also = 1/2
so probability of rain happening on three days and not on two days = 1/2^5
= 1/32
Since we have 10 ways of choosing this combination, total probability = 10/32
= 5/16
No of ways of choosing three days out of five is 5C3 = 10
probability of rain on a given day = 1/2
so probability of rain not happening on a given day is also = 1/2
so probability of rain happening on three days and not on two days = 1/2^5
= 1/32
Since we have 10 ways of choosing this combination, total probability = 10/32
= 5/16
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