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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · 11.2 Probability

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.12\dfrac{1}{2}
2.0.7
3.38\dfrac{3}{8}
4.14\dfrac{1}{4}
5.825=0.32\dfrac{8}{25} = 0.32
6.23\dfrac{2}{3}
7.16\dfrac{1}{6}
8.1225\dfrac{12}{25}
9.12\dfrac{1}{2}
10.0.25
Silver
11.0.9
12.518\dfrac{5}{18}
13.59\dfrac{5}{9}
14.512\dfrac{5}{12}
15.0.12
16.0.4
17.12\dfrac{1}{2}; yes, mutually exclusive
18.0.784
19.34\dfrac{3}{4}
20.0.65
Gold
21.0.032
22.Not independent (P(A) P(B)=0.10≠0.12P(A)\,P(B) = 0.10 \neq 0.12)
23.38\dfrac{3}{8}
24.528\dfrac{5}{28}
25.35\dfrac{3}{5}
26.0.5
27.0.067
28.0.12
29.0.375
30.49\dfrac{4}{9}
Platinum
31.≈0.269\approx 0.269
32.1219≈0.632\dfrac{12}{19} \approx 0.632
33.12\dfrac{1}{2}
34.n=6n = 6
35.415\dfrac{4}{15}
36.P(A∪B)=0.66P(A \cup B) = 0.66; not independent
37.0.3087
38.25\dfrac{2}{5}
39.13\dfrac{1}{3}
40.(a) 8; (b) 25\dfrac{2}{5}

Pack B — Answers

Bronze
1.13\dfrac{1}{3}
2.35\dfrac{3}{5}
3.25\dfrac{2}{5}
4.12\dfrac{1}{2}
5.38=0.375\dfrac{3}{8} = 0.375
6.58\dfrac{5}{8}
7.536\dfrac{5}{36}
8.425\dfrac{4}{25}
9.313\dfrac{3}{13}
10.0.30
Silver
11.0.75
12.512\dfrac{5}{12}
13.47\dfrac{4}{7}
14.25\dfrac{2}{5}
15.0.3
16.0.3
17.34\dfrac{3}{4}; yes, mutually exclusive
18.0.7599
19.23\dfrac{2}{3}
20.0.55
Gold
21.0.039
22.Independent (P(A) P(B)=0.15=P(A∩B)P(A)\,P(B) = 0.15 = P(A \cap B))
23.38\dfrac{3}{8}
24.16\dfrac{1}{6}
25.25\dfrac{2}{5}
26.0.4
27.0.0643
28.0.10
29.713≈0.538\dfrac{7}{13} \approx 0.538
30.2764\dfrac{27}{64}
Platinum
31.≈0.397\approx 0.397
32.58=0.625\dfrac{5}{8} = 0.625
33.59\dfrac{5}{9}
34.n=6n = 6
35.518\dfrac{5}{18}
36.P(A∪B)=0.9P(A \cup B) = 0.9; not independent
37.0.3456
38.120\dfrac{1}{20}
39.13\dfrac{1}{3}
40.(a) 10; (b) 25\dfrac{2}{5}

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) P(A)=12P(A) = \frac{1}{2}, P(B)=12P(B) = \frac{1}{2}, P(A∩B)=13P(A \cap B) = \frac{1}{3}, P(A∪B)=23P(A \cup B) = \frac{2}{3}. (b) Not independent: P(A)P(B)=14≠13P(A)P(B) = \frac{1}{4} \neq \frac{1}{3}.
Full working
A={2,4,6}A = \{2,4,6\}, B={4,5,6}B = \{4,5,6\}, A∩B={4,6}A \cap B = \{4,6\}, A∪B={2,4,5,6}A \cup B = \{2,4,5,6\}. So P(A)=12P(A) = \frac{1}{2}, P(B)=12P(B) = \frac{1}{2}, P(A∩B)=13P(A \cap B) = \frac{1}{3}, P(A∪B)=23P(A \cup B) = \frac{2}{3}. Independence test: 12⋅12=14≠13\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \neq \frac{1}{3}, so not independent.
2Problem 2
Answer
(a) Each branch: P(R)=47P(R) = \frac{4}{7}, P(B)=37P(B) = \frac{3}{7}. (b) 1649\frac{16}{49}. (c) 2449\frac{24}{49}.
Full working
With replacement the probabilities reset. (b) P(RR)=(47)2=1649P(RR) = \left(\frac{4}{7}\right)^2 = \frac{16}{49}. (c) P(RB)+P(BR)=2⋅47⋅37=2449P(RB) + P(BR) = 2 \cdot \frac{4}{7} \cdot \frac{3}{7} = \frac{24}{49}.
3Problem 3
Answer
(a) Branches scale on the second pick. (b) 16\frac{1}{6}. (c) 59\frac{5}{9}.
Full working
(b) P(BB)=49×38=16P(BB) = \frac{4}{9} \times \frac{3}{8} = \frac{1}{6}. (c) P(RB)+P(BR)=59⋅48+49⋅58=59P(RB) + P(BR) = \frac{5}{9}\cdot\frac{4}{8} + \frac{4}{9}\cdot\frac{5}{8} = \frac{5}{9}.
4Problem 4
Answer
(a) See working. (b) 0.067. (c) P(C∣T+)≈0.269P(C \mid T^+) \approx 0.269.
Full working
(b) P(T+)=0.02⋅0.9+0.98⋅0.05=0.018+0.049=0.067P(T^+) = 0.02 \cdot 0.9 + 0.98 \cdot 0.05 = 0.018 + 0.049 = 0.067. (c) P(C∣T+)=0.0180.067≈0.269P(C \mid T^+) = \frac{0.018}{0.067} \approx 0.269. Despite the test detecting 90% of true cases, only ~27% of positives are real — because the prevalence is low, false positives dominate.
5Problem 5
Answer
(a) 0.12. (b) 0.53. (c) Not independent.
Full working
(a) P(F∩S)=P(F) P(S∣F)=0.40×0.30=0.12P(F \cap S) = P(F)\,P(S \mid F) = 0.40 \times 0.30 = 0.12. (b) P(F∪S)=0.40+0.25−0.12=0.53P(F \cup S) = 0.40 + 0.25 - 0.12 = 0.53. (c) P(F)P(S)=0.40×0.25=0.10≠0.12P(F)P(S) = 0.40 \times 0.25 = 0.10 \neq 0.12, so not independent.
6Problem 6
Answer
(a) 0.032. (b) 0.625. (c) Wrong — 62.5% of defectives come from M2M_2.
Full working
(a) P(D)=0.6⋅0.02+0.4⋅0.05=0.032P(D) = 0.6 \cdot 0.02 + 0.4 \cdot 0.05 = 0.032. (b) P(M2∣D)=0.0200.032=0.625P(M_2 \mid D) = \frac{0.020}{0.032} = 0.625. (c) Even though M1M_1 produces more items, M2M_2's defect rate is over twice M1M_1's, so 62.5% of defectives come from M2M_2.
7Problem 7
Answer
(a) 0.343. (b) n=6n = 6. (c) 0.3087.
Full working
(a) 0.73=0.3430.7^3 = 0.343. (b) 1−0.3n≥0.999⇒0.3n≤0.001⇒n≥5.741 - 0.3^n \geq 0.999 \Rightarrow 0.3^n \leq 0.001 \Rightarrow n \geq 5.74. So n=6n = 6. (c) (53)(0.7)3(0.3)2=10×0.343×0.09=0.3087\binom{5}{3}(0.7)^3(0.3)^2 = 10 \times 0.343 \times 0.09 = 0.3087.
8Problem 8
Answer
(a) 120. (b) 25\frac{2}{5}. (c) 120\frac{1}{20}.
Full working
(a) 5×4×3×2=1205 \times 4 \times 3 \times 2 = 120. (b) Last digit even: 2 choices (2 or 4); first three from remaining 4 digits: 4⋅3⋅2=244\cdot 3\cdot 2 = 24. Favourable: 48. P=48120=25P = \frac{48}{120} = \frac{2}{5}. (c) Fix first = 1, last = 5; middle two from {2,3,4}\{2,3,4\}: 6 codes. P=6120=120P = \frac{6}{120} = \frac{1}{20}.
9Problem 9
Answer
(a) Tea only 18; both 12; coffee only 13; neither 7. (b) 3050,2550,1250,750\frac{30}{50}, \frac{25}{50}, \frac{12}{50}, \frac{7}{50}. (c) P(T∣C)=1225P(T \mid C) = \frac{12}{25}. (d) Not independent.
Full working
(a) Tea only =30−12=18= 30 - 12 = 18. Coffee only =25−12=13= 25 - 12 = 13. At least one =43= 43; neither =7= 7. (b) Divide each by 50. (c) n(T∩C)n(C)=1225\frac{n(T \cap C)}{n(C)} = \frac{12}{25}. (d) P(T)P(C)=3050⋅2550=7502500=0.3P(T)P(C) = \frac{30}{50}\cdot\frac{25}{50} = \frac{750}{2500} = 0.3. P(T∩C)=0.24P(T \cap C) = 0.24. 0.3≠0.240.3 \neq 0.24 → not independent.
10Problem 10
Answer
(a) 10%. (b) 15\frac{1}{5}. (c) Not independent. (d) Both 28%, neither 18%.
Full working
(a) P(S∪M)=1P(S \cup M) = 1 (everyone), so P(S∩M)=0.6+0.5−1=0.1P(S \cap M) = 0.6 + 0.5 - 1 = 0.1. (b) P(S∣M)=0.10.5=0.2P(S \mid M) = \frac{0.1}{0.5} = 0.2. (c) P(S)P(M)=0.30≠0.10P(S)P(M) = 0.30 \neq 0.10 → not independent. (d) Independent: P(S∩M)=0.28P(S \cap M) = 0.28; P(S∪M)=0.82P(S \cup M) = 0.82; neither =0.18= 0.18.
11Problem 11
Answer
(a) 34,25,310\frac{3}{4}, \frac{2}{5}, \frac{3}{10}. (b) Yes: 34⋅25=310\frac{3}{4} \cdot \frac{2}{5} = \frac{3}{10} ✓. (c) 34\frac{3}{4}.
Full working
(a) P(P)=150200=34P(P) = \frac{150}{200} = \frac{3}{4}. P(T)=80200=25P(T) = \frac{80}{200} = \frac{2}{5}. P(P∩T)=60200=310P(P \cap T) = \frac{60}{200} = \frac{3}{10}. (b) P(P) P(T)=34⋅25=620=310P(P)\,P(T) = \frac{3}{4} \cdot \frac{2}{5} = \frac{6}{20} = \frac{3}{10}. Equal to P(P∩T)P(P \cap T), so independent. (c) P(P∣T)=6080=34P(P \mid T) = \frac{60}{80} = \frac{3}{4} (same as P(P)P(P) — consistent with independence).
12Problem 12
Answer
(a) 0.022. (b) 1522≈0.682\frac{15}{22} \approx 0.682. (c) Wrong — Y's defect rate is 5× X's, so 68% of defectives come from Y.
Full working
(a) P(D)=0.7⋅0.01+0.3⋅0.05=0.007+0.015=0.022P(D) = 0.7 \cdot 0.01 + 0.3 \cdot 0.05 = 0.007 + 0.015 = 0.022. (b) P(Y∣D)=0.0150.022=1522≈0.682P(Y \mid D) = \frac{0.015}{0.022} = \frac{15}{22} \approx 0.682. (c) Although X produces more items, line Y is far more defect-prone, so a defective item is much more likely to come from Y. Volume alone is misleading without per-line defect rates.