(conditional probability bonus question) |
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P(F/E) = P(E and F)/P(E) = 0.2/0.6 = 1/3 | P(F/E) = P(E and F)/P(E) = 0.2/0.6 = 1/3 | ||
| − | + | ---- | |
| + | ==Comments/questions/discussions== | ||
| + | *Did you forget to put a link to your page on the assignment page? | ||
| + | *This would be a VERY easy exam problem. In a typical problem, you should expect to have to compute the probabilities for each event and for the intersection on your own, at the very least. -pm | ||
[[Bonus_point_1_ECE302_Spring2012_Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] | [[Bonus_point_1_ECE302_Spring2012_Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] | ||
Latest revision as of 07:10, 28 January 2013
Given that E and F are events such that P(E)=0.6 and P(F) = 0.3 and P(E and F) = 0.2 find P(E/F) and P(F/E)
solution:
P(E/F) = P(E and F)/P(F) = 0.2/0.3 = 2/3
P(F/E) = P(E and F)/P(E) = 0.2/0.6 = 1/3
Comments/questions/discussions
- Did you forget to put a link to your page on the assignment page?
- This would be a VERY easy exam problem. In a typical problem, you should expect to have to compute the probabilities for each event and for the intersection on your own, at the very least. -pm
