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## #1 2016-05-04 22:26:13

salem_ohio
Member
Registered: 2016-03-11
Posts: 37

### Weather forecast

There are two weather stations, station A and station B which are independent of each other. On average, the weather forecast accuracy of station A is 80% and that of station B is 90%. Station A predicts that there will be no rain tomorrow, whereas station B predicts rain. What is the probability that it rains tomorrow? We are not asking for the exact probability; we are just asking whether it is more likely to rain or not.

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## #2 2016-05-04 22:44:28

Relentless
Member
Registered: 2015-12-15
Posts: 624

### Re: Weather forecast

I think the chance of rain is

In any case, be inclined to expect rain.

Last edited by Relentless (2016-05-04 22:52:51)

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## #3 2016-05-04 23:28:39

salem_ohio
Member
Registered: 2016-03-11
Posts: 37

### Re: Weather forecast

I'll make sure to take my umbrella with me

Relentless wrote:

I think the chance of rain is

In any case, be inclined to expect rain.

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## #4 2016-05-04 23:53:42

Relentless
Member
Registered: 2015-12-15
Posts: 624

### Re: Weather forecast

Hi Typically, to find the chance of two random events both occurring, multiply the probabilities. In this case, however, they cannot both be right (.8*.9=.72) or both be wrong (.2*.1=.02). So we have to eliminate this artificial 74% to get probabilities out of the remaining 0.26 instead of 1.

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## #5 2016-05-05 05:36:01

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Weather forecast

Hi relentless;

Supposing we change the problem to Station A is right 50% of the time and Station B is right 50% of the time everything else stays the same. What do we get now?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #6 2016-05-05 05:45:46

Relentless
Member
Registered: 2015-12-15
Posts: 624

### Re: Weather forecast

Hey bobbym (:

We must eliminate the apparent chance they are both right or both wrong (25% each for total of 50%). Then:

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## #7 2016-05-05 05:55:16

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Weather forecast

Hi;

Okay, thanks.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #8 2016-05-06 02:58:53

salem_ohio
Member
Registered: 2016-03-11
Posts: 37

### Re: Weather forecast

My approach - but not sure if it is correct...
We must discern the following 4 cases for A and B:
1) They make the same forecast and both are right: Probability P = 4/5*8/9 = 32/45
2) Same forecast and both are wrong: - P = 1/5*1/9 = 1/45
3) Different forecasts and A is right: P =  4/5*1/9 = 4/45
4) Different forecasts and B is right: P = 1/5*8/9 = 8/45

In our case, we have A and B give different forecasts. In such cases, A is right with overall probability 4/45, while B is right with overall probability 8/45, that is double (that is, A: 1/3 while B: 2/3).
We are asking for the probability that it rains tomorrow, that is, the probability that B is right. I would therefore say that this is 2/3.
What do you think?

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## #9 2016-05-06 04:46:53

Member
Registered: 2016-04-16
Posts: 1,086

### Re: Weather forecast

salem_ohio wrote:

My approach - but not sure if it is correct...
We must discern the following 4 cases for A and B:
1) They make the same forecast and both are right: Probability P = 4/5*8/9 = 32/45
2) Same forecast and both are wrong: - P = 1/5*1/9 = 1/45
3) Different forecasts and A is right: P =  4/5*1/9 = 4/45
4) Different forecasts and B is right: P = 1/5*8/9 = 8/45

In our case, we have A and B give different forecasts. In such cases, A is right with overall probability 4/45, while B is right with overall probability 8/45, that is double (that is, A: 1/3 while B: 2/3).
We are asking for the probability that it rains tomorrow, that is, the probability that B is right. I would therefore say that this is 2/3.
What do you think?

You are telling same thing as Relentless but have put the value wrong.
3) must be 4/5*1/10 and
4) 1/5 *9/10

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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## #10 2016-05-06 04:55:10

Relentless
Member
Registered: 2015-12-15
Posts: 624

### Re: Weather forecast

Hi salem_ohio

You are right about the four cases and you understand the method But I agree with thickhead that you happened to put the wrong fractions of station B's probabilities: The denominator is 10. When you fix that, you get:
1) 18/25
2) 1/50
3) 2/25
4) 9/50

And also:
A) 4/13
B) 9/13

What you calculated, which happened to result in a 1/3:2/3 split, was for:
-Station A is right 80%, but
-Station B is right 88.8888...%

Last edited by Relentless (2016-05-06 04:58:06)

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## #11 2016-05-06 10:50:13

salem_ohio
Member
Registered: 2016-03-11
Posts: 37

### Re: Weather forecast

Ooops yes you are right, I did some mistakes in substituting the figures! Thanks

salem_ohio wrote:

My approach - but not sure if it is correct...
We must discern the following 4 cases for A and B:
1) They make the same forecast and both are right: Probability P = 4/5*8/9 = 32/45
2) Same forecast and both are wrong: - P = 1/5*1/9 = 1/45
3) Different forecasts and A is right: P =  4/5*1/9 = 4/45
4) Different forecasts and B is right: P = 1/5*8/9 = 8/45

In our case, we have A and B give different forecasts. In such cases, A is right with overall probability 4/45, while B is right with overall probability 8/45, that is double (that is, A: 1/3 while B: 2/3).
We are asking for the probability that it rains tomorrow, that is, the probability that B is right. I would therefore say that this is 2/3.
What do you think?

You are telling same thing as Relentless but have put the value wrong.
3) must be 4/5*1/10 and
4) 1/5 *9/10

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