Maths Question
17 April 2002, 08:13 AM
#1
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My maths is a bit rusty, can anyone help?
If A has a 90% chance of happening, B has a 80% chance of happening and C, 70%
Then the chances of A AND B happening are 90%*80% = 72%
and likewise for A AND C, 90%*70% = 63%
However what are the chances of A happening AND (either B OR C)?
What is the maths behind that one?
If A has a 90% chance of happening, B has a 80% chance of happening and C, 70%
Then the chances of A AND B happening are 90%*80% = 72%
and likewise for A AND C, 90%*70% = 63%
However what are the chances of A happening AND (either B OR C)?
What is the maths behind that one?
17 April 2002, 10:21 AM
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Ouch, mine's rusty too. Something to do with Bayes Theorem, I think.
Anyway, here's an 'interesting' page on probability that might help:
http://www.stats.gla.ac.uk/steps/glossary/probability.html
Anyway, here's an 'interesting' page on probability that might help:
http://www.stats.gla.ac.uk/steps/glossary/probability.html
17 April 2002, 11:17 AM
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You have to think about it backwards: there are four combinations of B and C
1) B and C both happen
2) B happens but C doesn't
3) C happens but B doesn't
4) Neither B nor C happen.
You want to capture the first three cases (I assume by 'or' you mean 'OR' not 'XOR'), so the probability of that is 100% minus the probability that neither happen, which is 1-(.2*.3)=.94 or 94%. Then multiply by the probability of A (90%) to get A AND (B OR C) as 84.6%.
You could also do the same by summing the first three cases, but it takes longer that way.
[Edited by carl - 4/17/2002 11:19:40 AM]
1) B and C both happen
2) B happens but C doesn't
3) C happens but B doesn't
4) Neither B nor C happen.
You want to capture the first three cases (I assume by 'or' you mean 'OR' not 'XOR'), so the probability of that is 100% minus the probability that neither happen, which is 1-(.2*.3)=.94 or 94%. Then multiply by the probability of A (90%) to get A AND (B OR C) as 84.6%.
You could also do the same by summing the first three cases, but it takes longer that way.
[Edited by carl - 4/17/2002 11:19:40 AM]
17 April 2002, 01:29 PM
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that doesn't sound right to me.
if a and b have a 70% chance of happening
A and C have a 60% chance of happening how can the chances of either of these happeing together be MORE than the chance of one happening on it's own?
if a and b have a 70% chance of happening
A and C have a 60% chance of happening how can the chances of either of these happeing together be MORE than the chance of one happening on it's own?
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17 April 2002, 07:15 PM
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disagree - it says A and (B or C)
i.e. A and B and C would not count
therefore it's 90%x(24%+14%) = 34.2%
(B not C is 80% x 30% = 24%)
(C not B is 70% x 20% = 14%)
(generally in questions like this they're kept simple hence XOR is never used, they would have put 'and or' instead of 'or' if both B and C could apply in addition to A)
isn't maths fun!
Gordo
[Edited by Gordo - 4/17/2002 7:17:31 PM]
i.e. A and B and C would not count
therefore it's 90%x(24%+14%) = 34.2%
(B not C is 80% x 30% = 24%)
(C not B is 70% x 20% = 14%)
(generally in questions like this they're kept simple hence XOR is never used, they would have put 'and or' instead of 'or' if both B and C could apply in addition to A)
isn't maths fun!
Gordo
[Edited by Gordo - 4/17/2002 7:17:31 PM]
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