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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. | ||
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| + | --[[User:Jrhaynie|Jrhaynie]] 14:14, 4 March 2009 (UTC) | ||
Revision as of 09:14, 4 March 2009
Back[[Category:MA375Spring2009Walther]
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)
