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Suppose there is string of eight bits, e.g.:

00100110

Bits are randomly chosen from the string. All choices are done equally likely.

Probability of choosing $0$: $p_0 = \frac{5}{8} = 0.625$

Prob. of choosing $1$: $p_1 = \frac{3}{8} = 0.375$

Suppose you have already chosen $0$ or $1$. Probability of choosing opposite char, and then again opposite char, is given with: $p(0 \wedge 1) = p_0 p_1 = 0.234$.

Without the "you have already chosen $0$ or $1$ ...", the probability would be: $p(0\wedge 1) = 2p_0 p_1 = 0.468$

Correct?

  • 5
    I have no idea what's going on...2012-04-26
  • 0
    Are you choosing one bit from the 8 given bits at random?2012-04-26
  • 0
    I have updated the question.2012-04-27

1 Answers 1