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

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

IB DP Maths AA: HL Papers 2: All Topics

Question

Nanda is responsible for entering survey data into a digital database. It is assumed that the accuracy of each entry is independent of any other.
Historical data indicates that Nanda has an error rate of 8% when entering surveys.
(a) Suppose Nanda processes 50 surveys on a given day.
(i) Calculate the probability that Nanda makes at most six errors.
(ii) If it is known that she made at most six errors, find the probability that she made exactly four errors.
On a different occasion, Nanda processes \( n \) surveys. The probability that she makes at most six errors on this day is approximately 0.367.
(b) Determine the value of \( n \).
Two other employees, Bryce and Carmen, also perform data entry for the same database. Their respective error rates are 6% and 11%. All entries remain independent.
The total workload is distributed such that Nanda enters 55%, Bryce enters 25%, and Carmen enters 20% of the surveys.
(c) Find the probability that a survey chosen at random was:
(i) entered with an error;
(ii) processed by Nanda, given that an error was found.
The following year, the error rates for Nanda and Bryce remained unchanged, as did the workload distribution. However, Carmen’s performance improved, and her new error rate is \( x\% \).
The probability that a randomly chosen survey contains an error is now equal to the probability of Carmen making an error.
(d) Calculate the value of \( x \).

Most-appropriate topic codes (IB Mathematics: Analysis and Approaches HL):

• SL 4.8: Binomial distribution — parts (a), (b)
• SL 4.11: Conditional probability — part (a)(ii)
• AHL 4.13: Bayes’ theorem — part (c)(ii)
• SL 4.6: Combined events, including mutually exclusive and independent events — parts (c), (d)
▶️ Answer/Explanation

(a)(i)
Let \( E \) be the number of inaccurate surveys entered by Nanda.
\( E \sim \mathrm{B}(50, 0.08) \)
\( P(E \leq 6) = 0.898128\ldots \)
\( \boxed{0.898} \) (accept 89.8%)

(a)(ii)
Conditional probability
\( P(E = 4 \mid E \leq 6) = \frac{P(E = 4)}{P(E \leq 6)} \)
\( P(E = 4) = 0.203654\ldots \)
\( P(E = 4 \mid E \leq 6) = \frac{0.203654\ldots}{0.898128\ldots} = 0.2267\ldots \)
\( \boxed{0.227} \) (accept 22.7%)

(b)
\( E \sim \mathrm{B}(n, 0.08) \), with \( P(E \leq 6) \approx 0.367 \)
Using tables/trial: for \( n = 94 \), \( P(E \leq 6) = 0.366746\ldots \approx 0.367 \).
\( \boxed{94} \)

(c)(i)
Use total probability
\( P(\text{inaccurate}) = 0.55 \times 0.08 + 0.25 \times 0.06 + 0.20 \times 0.11 \)
\( = 0.044 + 0.015 + 0.022 = 0.081 \)
\( \boxed{0.081} \) (accept 8.1%)

(c)(ii)
Bayes’ theorem
\( P(\text{Nanda} \mid \text{inaccurate}) = \frac{P(\text{Nanda} \cap \text{inaccurate})}{P(\text{inaccurate})} \)
\( = \frac{0.55 \times 0.08}{0.081} = \frac{0.044}{0.081} = 0.543209\ldots \)
\( \boxed{0.543} \) (accept 54.3%)

(d)
Let Carmen’s new error probability be \( \frac{x}{100} \).
Overall probability of an inaccurate survey
\( P(I) = 0.55 \times 0.08 + 0.25 \times 0.06 + 0.20 \times \frac{x}{100} \)
Given \( P(I) = \frac{x}{100} \):
\( \frac{x}{100} = 0.044 + 0.015 + 0.002x \)
\( \frac{x}{100} = 0.059 + 0.002x \)
Multiply by 100: \( x = 5.9 + 0.2x \)
\( 0.8x = 5.9 \)
\( x = 7.375 \)
\( \boxed{7.38} \) (accept 7.38%)

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