| Line 11: | Line 11: | ||
111 | 111 | ||
| − | So, the probability that there will be and odd number of | + | 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 | + | 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. |
Revision as of 10:12, 4 March 2009
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.
