OMPT probability basics practice: lesson and 10 problems

Answer: 0.11

Out of $36$ equally likely outcomes, the sum 9 arises from $(3,6),(4,5),(5,4),(6,3)$ — four of them. So the probability is $\frac{4}{36}=\frac{1}{9}\approx 0.11$.

Answer: B — $0.7$

Independence gives $P(A\cap B)=0.4\cdot 0.5=0.2$. Then the addition rule: $P(A\cup B)=0.4+0.5-0.2=0.7$.

Answer: 0.36

I chain the draws: the first is red with probability $\frac{5}{8}$; given that, the second is red with probability $\frac{4}{7}$. Multiplying: $\frac{5}{8}\cdot \frac{4}{7}=\frac{20}{56}=\frac{5}{14}\approx 0.36$.

Answer: B — $\frac{91}{216}$

"At least one" is easiest through the complement: the probability of no six at all is ${\left(\frac{5}{6}\right)}^3=\frac{125}{216}$. Subtracting from 1 gives $\frac{91}{216}\approx 0.42$.

Answer: Step 2

Step 1 frames the problem correctly. Step 2 adds the probabilities as if the subjects were mutually exclusive, but 25% of students sit in both classes and get counted twice. The addition rule fixes it: $0.60+0.45-0.25=0.80$. Step 3 is a patch over a broken calculation — getting a "probability" above 1 was the signal to go back, not to clip.

Answer: 0.16

I build a tree over the whole population. True positives: $0.02\cdot 0.95=0.019$. False positives: $0.98\cdot 0.10=0.098$. All positives: $0.019+0.098=0.117$. The conditional probability is $\frac{0.019}{0.117}\approx 0.16$. Rare conditions make even good tests unreliable on a single positive — the false positives from the huge healthy group swamp the true ones.

Answer: A — $\frac{5}{12}$

The condition "owns a car" shrinks the sample space to the 120 car owners. Of those, 50 also own a bike, so $P(\text{bike}\mid \text{car})=\frac{50}{120}=\frac{5}{12}$. The formula agrees: $\frac{P(\text{both})}{P(\text{car})}=\frac{50/200}{120/200}=\frac{50}{120}$.

Answer: 0.52

I draw the tree. Faulty and from P: $0.7\cdot 0.04=0.028$. Faulty and from Q: $0.3\cdot 0.10=0.030$. So $P(\text{faulty})=0.028+0.030=0.058$. Conditioning on faulty: $P(Q\mid \text{faulty})=\frac{0.030}{0.058}\approx 0.517$, which rounds to $0.52$. Q delivers under a third of the parts but over half of the faulty ones, because its fault rate is two and a half times P's.

Answer: 0.196

Two branches lead to a match. All black: $\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}=\frac{60}{336}$. All white: $\frac{3}{8}\cdot \frac{2}{7}\cdot \frac{1}{6}=\frac{6}{336}$. Adding the two branches: $\frac{66}{336}=\frac{11}{56}\approx 0.196$. Counting check: $\left(\genfrac{}{}{0ex}{}{5}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{3}\right)=10+1=11$ favourable sets out of $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$.

Answer: 0.5

I call $P(B)=p$. Independence gives $P(A\cap B)=0.3p$. The addition rule then reads $0.3+p-0.3p=0.65$, so $0.7p=0.35$ and $p=0.5$. Check: $0.3+0.5-0.15=0.65$. The whole question is two rules written down and one line of algebra.