A computer generates binary digits; P(0 at even place)=12, P(0 at odd place)=13.

Question

A computer generates binary digits; $P(0	ext{ at even place})=\dfrac12$, $P(0	ext{ at odd place})=\dfrac13$. Find the probability that "10" is followed by "01".

Accepted Answer

$P(1\text{ even})=\dfrac12$, $P(1\text{ odd})=\dfrac23$.

$$P = \frac23\cdot\frac12\cdot\frac13\cdot\frac12+\frac12\cdot\frac13\cdot\frac12\cdot\frac23 = \frac{1}{18}+\frac{1}{18}=\frac19$$

Solutions · 1