IB Maths AA HL Conditional Probability Questions

1. Identify the addition rule for probabilities: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

2. Substitute the known values into the equation to find the intersection: $0.85=0.45+0.60-P(A\cap B)$.

3. Solve for the probability of the intersection: $P(A\cap B)=1.05-0.85=0.20$.

4. State the formal definition of conditional probability: $P(A|B)=\frac{P(A\cap B)}{P(B)}$.

5. Substitute the calculated intersection and $P(B)$ into the conditional formula: $\frac{0.20}{0.60}$.

6. Simplify the fraction to find the exact probability: $\frac{\mathbf{1}}{\mathbf{3}}$.

1. Define the initial probabilities for the machines: $P(A)=0.60$ and $P(B)=0.40$.

2. Define the conditional probabilities for defects ($D$): $P(D|A)=0.03$ and $P(D|B)=0.05$.

3. Calculate the total probability of selecting a defective component using the law of total probability: $P(D)=P(A)P(D|A)+P(B)P(D|B)$.

4. Evaluate this total denominator probability: $P(D)=(0.60\times 0.03)+(0.40\times 0.05)=0.018+0.020=0.038$.

5. Apply Bayes' theorem to find the reverse conditional probability $P(A|D)$: $P(A|D)=\frac{P(A\cap D)}{P(D)}=\frac{P(A)P(D|A)}{P(D)}$.

6. Substitute the values and simplify: $\frac{0.018}{0.038}=\frac{18}{38}=\frac{9}{19}\approx \mathbf{0.474}$.

1. Write the general addition rule for probabilities: $P(C\cup D)=P(C)+P(D)-P(C\cap D)$.

2. Substitute the product of probabilities for the intersection, since the events are explicitly independent: $P(C\cap D)=P(C)P(D)$.

3. Set up the full algebraic equation by letting $x=P(D)$, meaning $P(C)=3x$: $0.68=3x+x-(3x)(x)$.

4. Rearrange the terms to form a standard quadratic equation: $3x^2-4x+0.68=0$.

5. Solve the quadratic equation by multiplying by $100$ and factorising (or using the quadratic formula): $300x^2-400x+68=0⟹75x^2-100x+17=0⟹x=0.2$ or $x=\frac{17}{15}$.

6. Reject the root that is strictly greater than $1$, giving the final exact probability: $\mathbf{P}\mathbf{(}\mathbf{D}\mathbf{)}\mathbf{=}\mathbf{0.2}$.