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jakehunter46
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Word Problems, Gde 5, 3d Ed. Medium Question Set #15 p 172

by jakehunter46 Sun Aug 26, 2012 5:35 pm

The probability of rainfall in City X on any given day is 30%. The probability of rainfall on any given day is independent of whether it rains on any other day.

Quantity A: The probability of rainfall in City X on at least one day out of two days.

Quantity B: The probability of no rainfall in City X on either of those two days.

The solution suggests Quantity A is .3+.3-.09=.51, while Quantity B is .7*.7=.49.

My question with quantity A is why are we removing the probability that it rains both days (.09)? If it rains both days, it still rains at least 1 day. Raining both days is raining at least 1 day. If quantity A has said the "probability of rain on one day" as opposed to "at least one day" then this is clearly p(a)+p(b)-p(ab), where I am going off track?

Help!

Jake
jgabry
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Re: Word Problems, Gde 5, 3d Ed. Medium Question Set #15 p 172

by jgabry Sun Aug 26, 2012 7:38 pm

Hey Jake,

The reason we subtract the .09 (which, as you say, is P(A and B)), is because we've counted it twice. So when we subtract it once, we're not removing it from our calculation, we're just making sure we don't double count it. I think the best way to grasp this intuitively is by seeing it visually with a Venn-diagram:

Image

In the image above, events A and B both happening would be the overlapping area. But, if you were to separating the regions, then would we have 2 full circles or 1 full circle and another circle with the shape of the overlap cut out of it?

If we don't cut the overlap out of either Event A or Event B then it will be counted twice.

In other words, P(A) + P(B) already contains P(A and B) twice by default, so we subtract it once to compensate for that.

I hope that made some sense and didn't make it more confusing!
nareshchowdary28
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Re: Word Problems, Gde 5, 3d Ed. Medium Question Set #15 p 172

by nareshchowdary28 Fri Aug 31, 2012 8:47 am

Yeah really confusing some times.

it goes like below but still I have a doubt on this

probability of rain fall on a day is 30/100
so no rain fall probability is 70/100

now Probability of rain fall on both the days is P(A)*P(B) = .3*.3 = .09
Probability of rain fall on no day is .7*.7 = .49 (But why can't it be 1-rain fall on both the days i.e 1-.09 = .91)

Probability of at least one day is clear that we should calc only either one day but not all days so the mutual exclusive comes into picture so P(A) + P(B) - P(A&B) (this is clear to me)

Please help me to get the clear idea on 2nd part.

Guys Please correct me if my fundamentals are wrong.

Cheers,
Naresh
jgabry
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Re: Word Problems, Gde 5, 3d Ed. Medium Question Set #15 p 172

by jgabry Fri Aug 31, 2012 5:28 pm

Hi Naresh,

There are four different possibilities for the two days we are considering:

Case I: Day 1 = No Rain and Day 2 = No Rain
Case II: Day 1 = No Rain and Day 2 = Rain
Case III: Day 1 = Rain and Day 2 = No Rain
Case IV: Day 1 = Rain and Day 2 = Rain

If P(Rain) = 3/10 and P(No Rain) = 7/10, then we can conclude:

Case I: Day 1 = No Rain and Day 2 = No Rain
= P(No Rain) x P(No Rain) = (7/10)^2 = 49/100

Case II: Day 1 = No Rain and Day 2 = Rain
= P(No Rain) x P(Rain) = (7/10) x (3/10) = 21/100

Case III: Day 1 = Rain and Day 2 = No Rain
= P(Rain) x P(No Rain) = (3/10) x (7/10) = 21/100

Case IV: Day 1 = Rain and Day 2 = Rain
= P(Rain) X P(Rain) = (3/10)^2 = 9/100


With this information we can really see the relationship between the different possible combinations of these outcomes. For example,

•The probability that it rains at least 1 of the 2 days:

= the probability that it rains only day 1, only day 2 or both on day 1 and on day 2

= case II + case III + case IV

= 21/100 + 21/100 + 9/100 = 51/100

We can also see that this is the same as 1 minus the probability that it doesn't rain at all.

= 1 - case I

= 1 - 49/100 = 51/100


And, therefore, we can see that the probability that it doesn't rain at all is equal to 1 minus the probability that it rains on at least 1 day (not just 1 - the probability it rains both days):

= 1 - case II - case III - case IV = 1 - 21/100 - 21/100 - 9/100 = 49/100


To solve the original problem we didn't need to do all of this, but sometimes it can help to write it out and really see how everything fits together.

I hope that helps!
tommywallach
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Re: Word Problems, Gde 5, 3d Ed. Medium Question Set #15 p 172

by tommywallach Sat Sep 01, 2012 7:42 pm

Great explanation from JGabry!

Remember, when you have multiple separate events, an OUTCOME cannot be simply one event, but a pattern of all of them.

In other words, if you're worried about the weather on six days, the probability that it will only rain on Monday is not JUST the probability of rain on Monday, but the probability of rain on Monday multiplied by the probability of NO rain on all the other days.

Well done JG!

-t