IB Mathematics SL 4.6 Use of Venn diagrams, tree diagrams AI SL Paper 2- Exam Style Questions- New Syllabus
Question
A specific type of generator operates only if a designated switch is functional. The generator is equipped with a main switch, A, and a backup switch, B.
The manufacturer asserts that the likelihood of switch A failing within one month of installation is 0.1 and the probability of the less expensive switch B failing within one month is 0.3. The failure of one switch is independent of the other’s condition.
If both switches fail, the generator must shut down for switch replacement. Both switches are replaced after a month of use, regardless of failure, or whenever a shutdown is required.
The following tree diagram illustrates the probabilities of a switch failing within one month after replacement, assuming the manufacturer’s claim holds true.
Switch A → Fail (0.1) → Switch B: Fail = b, Not fail = c
Switch A → Not fail (a) → Switch B: Fail = b, Not fail = c
(a) Write down the values of
(i) \(a\)
(ii) \(b\)
(iii) \(c\)[2]
(b) Hence determine the probability that the generator needs to shut down within one month of the switches being replaced.[1]
The owner of the generator doubts the manufacturer’s claims and reviews records from the past 200 instances of switch replacements. The records indicate whether no switches, one switch, or two switches failed.
The data collected by the owner are presented in the following table.
| No switch fails | One switch fails | Two switches fail |
|---|---|---|
| 118 | 72 | 10 |
(c) Conduct a \(\chi^2\) goodness of fit test at the 5% significance level to evaluate whether the manufacturer’s claims are accurate using the following hypotheses:
\(H_0:\) The manufacturer’s claims are accurate;
\(H_1:\) The manufacturer’s claims are not both accurate.[9]
▶️ Answer/Explanation
(a) Using the manufacturer’s probabilities and independence:
(i) \( a = P(\text{A not fail in a month}) = 1 – 0.1 \); \( = 0.9 \)
(ii) \( b = P(\text{B fail in a month}) \); \( = 0.3 \) (independent of A)
(iii) \( c = P(\text{B not fail in a month}) = 1 – 0.3 \); \( = 0.7 \)
(b) The generator shuts down within the month only if both switches fail in that month:
\( P(\text{shutdown in month}) = P(\text{A fails and B fails}) \); \( = P(\text{A fails}) × P(\text{B fails}) \); \( = 0.1 × 0.3 \); \( = 0.03 \)
(c) Chi-square goodness-of-fit test (5% level)
Model probabilities under \( H_0 \) (manufacturer correct):
No switch fails: \( P(\bar A \& \bar B) = 0.9 × 0.7 \); \( = 0.63 \)
Exactly one fails: \( P(A \& \bar B) + P(\bar A \& B) \); \( = 0.1 × 0.7 + 0.9 × 0.3 \); \( = 0.07 + 0.27 \); \( = 0.34 \)
Two fail: \( P(A \& B) = 0.1 × 0.3 \); \( = 0.03 \)
Expected counts for \( n = 200 \):
\( E = (200 × 0.63, 200 × 0.34, 200 × 0.03) \); \( = (126, 68, 6) \)
Observed counts (from records): \( O = (118, 72, 10) \)
Test statistic:
\( \chi^2 = \sum \frac{(O – E)^2}{E} \); \( = \frac{(118 – 126)^2}{126} + \frac{(72 – 68)^2}{68} + \frac{(10 – 6)^2}{6} \); \( = \frac{64}{126} + \frac{16}{68} + \frac{16}{6} \); \( \approx 0.508 + 0.235 + 2.667 \); \( = 3.4099 \)
Degrees of freedom: \( k – 1 = 3 – 1 \); \( = 2 \)
p-value: \( P(\chi^2_{(2)} \ge 3.4099) \approx 0.182 \)
Decision & conclusion: Since \( p = 0.182 > 0.05 \); we do not reject \( H_0 \); There is insufficient evidence to contradict the manufacturer’s stated failure probabilities; the data are consistent with the claims.