Difference between revisions of "Conditional probability"
(New page: A '''conditional probability''' is the probability that some event A occurs given that a different event B has already occurred. This is written P(A|B) (often read as "the probability ...) |
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<math> | <math> | ||
| − | P(A|B)=P(AB) | + | P(A|B)=\frac{P(AB)}{P(B)} |
</math> | </math> | ||
where P(AB) is the probability of both events A and B occurring and P(B) is the probability of event B occurring. | where P(AB) is the probability of both events A and B occurring and P(B) is the probability of event B occurring. | ||
| − | + | ==Example== | |
A bag has 5 red balls and 3 green balls inside it. You pick out a ball and do not replace it in the bag afterwards. | A bag has 5 red balls and 3 green balls inside it. You pick out a ball and do not replace it in the bag afterwards. | ||
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A. There are now 4 red balls and 3 greens, so: 4/7 | A. There are now 4 red balls and 3 greens, so: 4/7 | ||
| − | In the second case, the question requires a conditional probability: P(red|red). | + | In the second case, the question requires a conditional probability: P(red on 2nd draw|red on 1st draw). |
| − | + | ==See also== | |
| − | [[Bayes | + | [[Bayes theorem]] |
[[Category:Probability and Statistics]] | [[Category:Probability and Statistics]] | ||
Latest revision as of 06:42, July 13, 2016
A conditional probability is the probability that some event A occurs given that a different event B has already occurred. This is written P(A|B) (often read as "the probability of A given B"). The conditional probability satisfies the equation
<math> P(A|B)=\frac{P(AB)}{P(B)} </math>
where P(AB) is the probability of both events A and B occurring and P(B) is the probability of event B occurring.
Example
A bag has 5 red balls and 3 green balls inside it. You pick out a ball and do not replace it in the bag afterwards.
Q. What is the probability of removing a red ball?
A. 5/8
Q. What is the probability of removing a red ball, given that you have already removed a red ball?
A. There are now 4 red balls and 3 greens, so: 4/7
In the second case, the question requires a conditional probability: P(red on 2nd draw|red on 1st draw).