Computing probabilities in practice

Is it likely that two randomly chosen students are both girls? How safe is a batch of products if a few are faulty? What are the odds of winning the lottery? These questions are all the same type: count the favourable outcomes, count all outcomes, divide. The difficulty is in the counting, and that is where permutations and combinations come in.

A strategy that works

1. Define the experiment and decide whether order matters. Drawing objects "at once" means order does not matter, so use combinations.
2. Count $n(\Omega)$ with the counting rules.
3. Count $n(A)$: split into cases or use the multiplication rule. If the event says "at least one", consider the complement.
4. Divide and simplify: $P(A)=\frac{n(A)}{n(\Omega)}$. Check that $0\le P\le 1$.

Balls from an urn

Choosing a committee

Quality control

Lotteries and cards

Key terms

Summary

• Strategy: define the experiment, count $n(\Omega)$, count $n(A)$, divide.
• Objects taken "at once" or "without order": combinations.
• "At least one": $P=1-P(\text{none})$.
• Always check that $0\le P\le1$ and simplify.

Practice

Show solutions

1. $\frac{\binom{6}{2}}{\binom{10}2}=\frac{15}{45}=\frac13$.
2. $\frac{6\cdot4}{45}=\frac{24}{45}=\frac8{15}$.
3. $\frac{\binom{12}3}{\binom{20}3}=\frac{220}{1140}=\frac{11}{57}$.
4. $1-\frac{11}{57}=\frac{46}{57}$.
5. $\frac{\binom{13}3}{\binom{15}3}=\frac{286}{455}=\frac{22}{35}$.
6. The product is odd only if both numbers are odd: $3\cdot3=9$ of $36$ outcomes. $P(\text{even})=1-\frac9{36}=\frac34$.
7. $\frac{\binom{13}3}{\binom{52}3}=\frac{286}{22\,100}=\frac{11}{850}$.
8. Three favourable tickets out of $\binom{45}{6}$ possible draws: $\frac3{8\,145\,060}=\frac1{2\,715\,020}$.
9. $n(\Omega)=\binom{9}{4}=126$. Favourable: $\binom{5}{2}\binom{4}{2}=10\cdot6=60$. $P=\frac{60}{126}=\frac{10}{21}$.
10. $n(\Omega)=6^3=216$. Sum $9$: there are $25$ ordered triples, so $P=\frac{25}{216}$. No six: $5^3=125$ triples, so $P(\text{at least one six})=1-\frac{125}{216}=\frac{91}{216}$.