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?