IB Mathematics SL 4.5 Concepts of trial, outcome, equally likely outcomes AI SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
\( \frac{1}{12} \)
Requires heads and die showing 6
\( P(7) = P(\text{heads}) \times P(\text{6}) \)
\( \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \approx 0.0833 \)
Result: \( \frac{1}{12} \) [2]
(b)
Possible points: 1 to 7
| Final number of points | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Probability | \(\frac{1}{12}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{12}\) |
Tails: points = die score (1 to 6)
Heads: points = die score + 1 (2 to 7)
For \( x = 1 \): \( P(1) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \)
For \( x = 2 \): \( P(2) = \left( \frac{1}{2} \times \frac{1}{6} \right) + \left( \frac{1}{2} \times \frac{1}{6} \right) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6} \)
Similarly for \( x = 3, 4, 5, 6 \)
For \( x = 7 \): \( P(7) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \)
Result: As shown in the table [3]
(c)
4
Expected value formula: \( E(X) = \sum x P(X = x) \)
\( E(X) = (1 \times \frac{1}{12}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6}) + (7 \times \frac{1}{12}) \)
\( \frac{1}{12} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} + \frac{7}{12} \)
\( \frac{1}{12} + \frac{7}{12} + \frac{2 + 3 + 4 + 5 + 6}{6} \)
\( \frac{8}{12} + \frac{20}{6} = \frac{2}{3} + \frac{10}{3} = 4 \)
Result: 4 [3]
▶️ Answer/Explanation
(a)
Total outcomes: \( 6 \times 6 = 36 \)
\( T = \text{maximum of two dice} \)
\( P(T=1) \): (1,1), count = 1, \( \frac{1}{36} \)
\( P(T=2) \): (2,2), (2,1), (1,2), count = 3, \( \frac{3}{36} \)
\( P(T=3) \): (3,3), (3,1), (1,3), (3,2), (2,3), count = 5, \( \frac{5}{36} \)
\( P(T=4) \): (4,4), (4,1), (1,4), (4,2), (2,4), (4,3), (3,4), count = 7, \( \frac{7}{36} \)
\( P(T=5) \): (5,5), (5,1), (1,5), (5,2), (2,5), (5,3), (3,5), (5,4), (4,5), count = 9, \( \frac{9}{36} \)
\( P(T=6) \): (6,6), (6,1), (1,6), (6,2), (2,6), (6,3), (3,6), (6,4), (4,6), (6,5), (5,6), count = 11, \( \frac{11}{36} \)
Result: As shown in the table [2]
(b)
(i) \( \frac{8}{9} \)
\( P(T \geq 3) = P(T=3) + P(T=4) + P(T=5) + P(T=6) \)
\( \frac{5}{36} + \frac{7}{36} + \frac{9}{36} + \frac{11}{36} = \frac{32}{36} = \frac{8}{9} \approx 0.888888 \)
Result: \( \frac{8}{9} \) [2]
(ii) \( \frac{11}{32} \)
Conditional probability: \( P(T=6 | T \geq 3) = \frac{P(T=6)}{P(T \geq 3)} \)
\( P(T=6) = \frac{11}{36} \), \( P(T \geq 3) = \frac{32}{36} \)
\( \frac{\frac{11}{36}}{\frac{32}{36}} = \frac{11}{32} \approx 0.34375 \)
Result: \( \frac{11}{32} \) [1]
(c)
\( \frac{161}{36} \)
Expected value formula: \( E(T) = \sum t P(T = t) \)
\( E(T) = (1 \times \frac{1}{36}) + (2 \times \frac{3}{36}) + (3 \times \frac{5}{36}) + (4 \times \frac{7}{36}) + (5 \times \frac{9}{36}) + (6 \times \frac{11}{36}) \)
\( = \frac{1 + 6 + 15 + 28 + 45 + 66}{36} = \frac{161}{36} \approx 4.472222 \)
Result: \( \frac{161}{36} \) [2]