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[[Category:MA375Spring2009Walther]] | [[Category:MA375Spring2009Walther]] | ||
| + | [[Category:MA375]] | ||
| + | [[Category:math]] | ||
| + | [[Category:discrete math]] | ||
| + | [[Category:problem solving]] | ||
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| + | =[[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 | ||
| + | ---- | ||
There are 8 options for bitstrings of length 3 : | There are 8 options for bitstrings of length 3 : | ||
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So, the probability that there will be and odd number of 1s occurs when there is either one 1 or three 1s making P(E)=1/2. | So, the probability that there will be and odd number of 1s occurs when there is either one 1 or three 1s making P(E)=1/2. | ||
The probability that the bitstring starts with zero, P(F), is also 1/2. Therefore, P(E)*P(F)=1/4. Because we know that the cases that there is an odd number of 1s and that starts with a zero is the intersection of the two, we see that P(EnF)=1/4. Since P(E)*P(F)=1/4=P(EnF)we can say that the events are indeed independent. | The probability that the bitstring starts with zero, P(F), is also 1/2. Therefore, P(E)*P(F)=1/4. Because we know that the cases that there is an odd number of 1s and that starts with a zero is the intersection of the two, we see that P(EnF)=1/4. Since P(E)*P(F)=1/4=P(EnF)we can say that the events are indeed independent. | ||
| + | |||
| + | --[[User:Jrhaynie|Jrhaynie]] 14:14, 4 March 2009 (UTC) | ||
| + | ---- | ||
| + | [[MA375_%28WaltherSpring2009%29|Back to MA375, Spring 2009, Prof. Walther]] | ||
Latest revision as of 09:21, 20 May 2013
MA375: Solution to a homework problem from this week or last week's homework
Spring 2009, Prof. Walther
There are 8 options for bitstrings of length 3 : 000 001 010 100 011 110 101 111
So, the probability that there will be and odd number of 1s occurs when there is either one 1 or three 1s making P(E)=1/2. The probability that the bitstring starts with zero, P(F), is also 1/2. Therefore, P(E)*P(F)=1/4. Because we know that the cases that there is an odd number of 1s and that starts with a zero is the intersection of the two, we see that P(EnF)=1/4. Since P(E)*P(F)=1/4=P(EnF)we can say that the events are indeed independent.
--Jrhaynie 14:14, 4 March 2009 (UTC)
