Question 11 of 12

Question 11

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A population of 1000 with the headache/fever/both counts tabulated above, and a random 8-bit byte with each bit independently 0 or 1 with probability 0.5.

Find. (a) $P(H\cup F)$, $P(H|F)$, $P(F|H)$, $P(H|\overline F)$; (b) $P(\text{no run of} \geq5 \text{ consecutive 0's in 8 bits})$.

Approach. (a) Apply inclusion-exclusion for the union and the conditional-probability definition $P(A|B)=P(A\cap B)/P(B)$ for each conditional; (b) enumerate/complement-count over all $2^8=256$ equally likely bytes for the maximum run of consecutive 0's.

1. 11a-i) $P(\text{sick})=P(H\cup F)$. By inclusion-exclusion: $$P(H\cup F)=P(H)+P(F)-P(H\cap F)=0.01+0.02-0.005=0.025$$ $\boxed{P(\text{sick})=0.025}$
2. 11a-ii) $P(H|F)$. $$P(H|F)=\frac{P(H\cap F)}{P(F)}=\frac{0.005}{0.02}=0.25$$ $\boxed{P(H|F)=0.25}$
3. 11a-iii) $P(F|H)$. $$P(F|H)=\frac{P(H\cap F)}{P(H)}=\frac{0.005}{0.01}=0.5$$ $\boxed{P(F|H)=0.5}$
4. 11a-iv) $P(H|\overline F)$. $P(H\cap\overline F)=P(H)-P(H\cap F)=0.01-0.005=0.005$ and $P(\overline F)=1-P(F)=0.98$, so $$P(H|\overline F)=\frac{0.005}{0.98}=\frac{1}{196}\approx0.00510$$ $\boxed{P(H|\overline F)=\dfrac{1}{196}\approx0.0051}$
5. 11b) No run of $\geq5$ consecutive 0's in a random byte. There are $2^8=256$ equally likely bytes. Enumerating all 256 and checking the longest run of consecutive 0's in each: exactly 20 of the 256 bytes contain a run of 5 or more zeros (e.g. 00000xxx and its shifted/complemented positions), leaving $$256-20=236 \text{ bytes with every zero-run} \leq 4$$ $$P(\text{no run}\geq5 \text{ zeros})=\frac{236}{256}=\frac{59}{64}\approx0.9219$$ $\boxed{P=\dfrac{59}{64}\approx0.922}$