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**salem_ohio****Member**- Registered: 2016-03-11
- Posts: 35

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

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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**salem_ohio****Member**- Registered: 2016-03-11
- Posts: 35

I'll make sure to take my umbrella with me

Solution please??

Relentless wrote:

I think the chance of rain is

In any case, be inclined to expect rain.

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

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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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,752

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

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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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,752

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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**salem_ohio****Member**- Registered: 2016-03-11
- Posts: 35

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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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,051

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 = 8/45In 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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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

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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**salem_ohio****Member**- Registered: 2016-03-11
- Posts: 35

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

thickhead wrote:

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 = 8/45In 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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