Probability: worked solution — ASRJC 2025 H2 Math Prelim Paper 2
Question
- In a survey, 100 library members were asked the number of days per month they visit the
library. The results are:
1 or less 2 to 4 5 or more Male 15 25 22 Female 20 \(18-n\) \(n\) Events are defined as:
- \(A\): the member visits on 4 days or fewer per month.
- \(B\): the member visits on 2 days or more per month.
- \(F\): the member is female.
- \(P(A \cup B')\)
- \(P(F \mid A')\)
- Given that \(P(A \cap B) = \dfrac{3}{10}\), find the value of \(n\). Hence determine if \(A\) and \(B\) are independent, justifying your answer.
- A librarian creates 4-letter access codes from the letters of the word BOOKKEEPER.
- How many access codes can be formed if all 4 letters are distinct?
- Find the total number of access codes with no restriction.
- In how many ways can the team be formed if at least one volunteer must be chosen from each age group?
Find the following probabilities in terms of \(n\).
The librarian selects a team of 12 volunteers from 18 (5 youths, 6 young adults, 7 seniors).
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(a)
Summary table (\(n\) denotes female members visiting 5 or more days):
| \(\leq 1\) day | 2 | ndash;4 days | \(\geq 5\) days | Total | |
|---|---|---|---|---|---|
| Male | 15 | 25 | 22 | 62 | |
| Female | 20 | \(18-n\) | \(n\) | 38 | |
| Total | 35 | \(43-n\) | \(n+22\) | 100 |
Events: \(A = \{\leq 4\text{ days}\}\), \(B = \{\geq 2\text{ days}\}\), \(F = \{\text{female}\}\).
(a)(i)
\[\begin{aligned} \mathrm{P}(A \cup B') &= \mathrm{P}(A) \\ &= \frac{15 + 25 + 20 + (18-n)}{100} \quad \text{or} \quad \frac{100 - 22 - n}{100} \\ &= \frac{78 - n}{100} \end{aligned}\] \[ P(A) = \frac{15+25+20+(18-n)}{100} = \frac{78-n}{100}. \]\(P(A \cup B') = \dfrac{78-n}{100}\).
(a)(ii) \(A' = \{\geq 5\text{ days}\}\). Members in \(A'\): 22 males + \(n\) females \(= n+22\).
\(P(F \cap A') = \dfrac{n}{100}\).
\[ P(F \mid A') = \frac{P(F\cap A')}{P(A')} = \frac{n/100}{(n+22)/100} = \frac{n}{n+22}. \]\(P(F \mid A') = \dfrac{n}{n+22}\).
(a)(iii)
Given \(\mathrm{P}(A \cap B) = \dfrac{3}{10}\)
\[\begin{aligned} \mathrm{P}(A \cap B) &= \frac{25 + 18 - n}{100} = \frac{43 - n}{100} \\[6pt] \frac{43 - n}{100} &= \frac{3}{10} \implies n = 13 \\[6pt] \mathrm{P}(A)\mathrm{P}(B) &= \frac{65}{100} \times \frac{65}{100} = \frac{169}{400} \neq \frac{3}{10} \end{aligned}\]Since \(\mathrm{P}(A)\mathrm{P}(B) \neq \mathrm{P}(A \cap B)\), \(A\) and \(B\) are NOT independent.
\(n = 13\). Since \(P(A)P(B) = \dfrac{169}{400} \neq \dfrac{3}{10} = P(A\cap B)\), events \(A\) and \(B\) are not independent.
(b) Letters of BOOKKEEPER: B, O, O, K, K, E, E, E, P, R. Distinct: B, O, K, E, P, R (6 types); repeats: 3E, 2O, 2K.
(b)(i) All 4 distinct: choose 4 from 6 distinct letters, arrange: \[ ^6C_4 \times 4! = 15 \times 24 = 360. \]
360 access codes.
(b)(ii)
Case 1: All distinct (from b(i))
Number of 4-letter code words \(= {}^6C_4 \times 4! = 360\)
Case 2: One pair of repeated letters. (Eg XXYZ)
Number of 4-letter code words \(= {}^3C_1 \times {}^5C_2 \times \dfrac{4!}{2!} = 360\)
Case 3: Two pairs of repeated letters. (Eg XXYY)
Number of 4-letter code words \(= {}^3C_2 \times \dfrac{4!}{2!\,2!} = 18\)
Case 4: One set of 3 repeated letters (Eg EEEX)
Number of 4-letter code words \(= {}^5C_1 \times \dfrac{4!}{3!} = 20\)
\(\therefore\) Total number of 4-letter code word \(= 360 + 360 + 18 + 20 = 758\)
Total \(= 360 + 360 + 18 + 20 = \mathbf{758}\) access codes.
(b)(iii) Total ways to choose 12 from 18: \(^{18}C_{12} = 18564\).
Subtract cases missing at least one group (use complement):
- No youth (choose from 13): \(^{13}C_{12} = 13\)
- No young adult (choose from 12): \(^{12}C_{12} = 1\)
- No senior (choose from 11): \(^{11}C_{12} = 0\) (impossible)
- No youth & no young adult (choose from 7): \(^7C_{12} = 0\)
- Other double/triple intersections: all 0.
Required \(= 18564 - 13 - 1 - 0 = \mathbf{18550}\) ways.