IBDP Maths AHL 4.13 Use of Bayes’ theorem for a maximum of three events AA HL Paper 2- Exam Style Questions- New Syllabus

(a)

Since each semiconductor is either defective or non-defective independently, the number of defective semiconductors follows a binomial distribution.

For Machine A:

\( P(X=2)=\binom{10}{2}(0.1)^2(0.9)^8 =0.193710\ldots \)

For Machine B:

\( P(X=2)=\binom{10}{2}(0.05)^2(0.95)^8 =0.0746347\ldots \)

Since each machine is equally likely to be selected,

\( P(\text{2 defective}) =\frac12(0.193710\ldots)+\frac12(0.0746347\ldots) \)

\( =0.134172\ldots \)

\( \boxed{P(\text{2 defective})\approx0.134} \)

The total probability is obtained by averaging the probabilities from both machines because each machine has probability \(\tfrac12\) of being chosen.

✅ Answer: \(0.134\)

(b)

Use conditional probability (Bayes’ theorem):

\( P(A\mid \text{2 defective}) = \frac{P(A)\times P(\text{2 defective}\mid A)} {P(\text{2 defective})} \)

Substitute the known values:

\( = \frac{0.5(0.193710\ldots)} {0.134172\ldots} \)

\( = 0.72187\ldots \)

\( \boxed{P(A\mid \text{2 defective})\approx0.722} \)

This means that, given exactly two defective semiconductors were observed, there is about a \(72.2\%\) chance they came from Machine A because Machine A has the higher defect rate.

✅ Answer: \(0.722\)