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| − | + | [[Category:MA375Spring2009Walther]] | |
| + | [[Category:MA375]] | ||
| + | [[Category:math]] | ||
| + | [[Category:discrete math]] | ||
| + | [[Category:problem solving]] | ||
| − | + | =[[MA375]]: [[MA_375_Spring_2009_Walther_Week_5| Solution to a homework problem from this week or last week's homework]]= | |
| + | Spring 2009, Prof. Walther | ||
| + | ---- | ||
| − | |||
| − | |||
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| − | (0.5)*(0.5)≠0 | + | |
| − | + | If Independent then P(H)*P(T)=P(H∩T) | |
| + | |||
| + | Sample Case: One flip of coin | ||
| + | |||
| + | P(H)=0.5 | ||
| + | P(T)=0.5 | ||
| + | P(H∩T)=0 (You can't have both H and T in one flip) | ||
| + | |||
| + | (0.5)*(0.5)≠0 | ||
| + | Not independent | ||
| + | |||
| + | ---- | ||
| + | [[MA375_%28WaltherSpring2009%29|Back to MA375, Spring 2009, Prof. Walther]] | ||
Latest revision as of 09:23, 20 May 2013
MA375: Solution to a homework problem from this week or last week's homework
Spring 2009, Prof. Walther
If Independent then P(H)*P(T)=P(H∩T)
Sample Case: One flip of coin
P(H)=0.5 P(T)=0.5 P(H∩T)=0 (You can't have both H and T in one flip)
(0.5)*(0.5)≠0 Not independent
