L31 Solutions — Introduction to Probability

1. Basic probability. Fluency

   • (a) 3 even numbers (2,4,6) out of 6. P = 1/2.
   • (b) P(red) = 5/12.
   • (c) P(A′) = 1 − 0.35 = 0.65.
   • (d) PROBABILITY: 11 letters, vowels O,A,I,I,Y — wait: P-R-O-B-A-B-I-L-I-T-Y. Vowels: O, A, I, I = 4 vowels (Y is sometimes vowel; count as consonant here). P = 4/11.

2. Mutually exclusive events. Fluency

   • (a) 0.3 + 0.45 = 0.75.
   • (b) 4/52 + 4/52 = 8/52 = 2/13.
   • (c) 1/6 + 1/6 = 1/3.
   • (d) P(A or B) = 2/7 + 3/7 = 5/7. P(neither) = 1 − 5/7 = 2/7.

3. Addition rule. Fluency

   • (a) 0.5 + 0.4 − 0.15 = 0.75.
   • (b) P(red) = 26/52, P(face) = 12/52, P(red face) = 6/52. P(red or face) = 26/52 + 12/52 − 6/52 = 32/52 = 8/13.
   • (c) P(A ∩ B) = 0.5 + 0.4 − 0.8 = 0.1.
   • (d) P(A ∪ B) = 0.6+0.5−0.3 = 0.8. P(A′ ∩ B′) = 1 − P(A∪B) = 1 − 0.8 = 0.2.

4. Sample spaces. Fluency

   • (a) S = {HH, HT, TH, TT}. P(exactly one H) = 2/4 = 1/2.
   • (b) 6 × 6 = 36 outcomes.
   • (c) P(spade) = 13/52, P(ace) = 4/52, P(ace of spades) = 1/52. P(spade or ace) = 13/52+4/52−1/52 = 16/52 = 4/13.
   • (d) Primes in 1–8: 2, 3, 5, 7. P = 4/8 = 1/2.

5. Venn diagram (Football and Basketball). Understanding

   • (a) P(F ∩ B) = 8/40 = 1/5.
   • (b) F only = 22−8 = 14. P = 14/40 = 7/20.
   • (c) P(F ∪ B) = (14+8+10)/40 = 32/40 = 4/5.
   • (d) Neither = 40−32 = 8. P = 8/40 = 1/5.

6. Two-way table (Coffee and Tea). Understanding

   • (a) 120/200 = 3/5 = 0.6.
   • (b) 50/200 = 1/4 = 0.25.
   • (c) (120+90−50)/200 = 160/200 = 4/5 = 0.8.
   • (d) No — 50 people drink both. P(Coffee ∩ Tea) = 0.25 ≠ 0. Mutually exclusive requires P(both) = 0.

7. Experimental probability. Understanding

   • (a) 310/500 = 0.62.
   • (b) 12/300 = 0.04 = 4%.
   • (c) It converges toward the theoretical probability (Law of Large Numbers).
   • (d) No — random variation is expected. 18 occurrences in 60 rolls is unusual but not conclusive. A much larger number of trials (e.g. 600 or 6000) would be needed to establish bias.

8. Tree diagram (3R 2B, without replacement). Understanding

   • (a) First draw: R (3/5) or B (2/5). If R first: R (2/4) or B (2/4). If B first: R (3/4) or B (1/4). Four branches: RR, RB, BR, BB.
   • (b) P(RR) = (3/5)×(2/4) = 6/20 = 3/10.
   • (c) P(RB) + P(BR) = (3/5)(2/4) + (2/5)(3/4) = 6/20 + 6/20 = 12/20 = 3/5.
   • (d) P(at least one B) = 1 − P(RR) = 1 − 3/10 = 7/10.

9. Card game (3 draws without replacement). Problem Solving

   • (a) P(AAA) = (4/52)×(3/51)×(2/50) = 24/132600 = 1/5525.
   • (b) P(first = Ace) = 4/52 = 1/13.
   • (c) P(no aces) = (48/52)×(47/51)×(46/50) = 103776/132600 ≈ 0.7826. P(at least one ace) = 1 − 0.7826 ≈ 0.217.
   • (d) P(all same suit) = 4 × (13/52)×(12/51)×(11/50) = 4×1716/132600 = 6864/132600 = 572/11050 ≈ 0.0518.

10. Faulty bulbs (P(def) = 0.03, 3 bulbs independent). Problem Solving

    • (a) 0.03.
    • (b) 0.03³ = 0.000027.
    • (c) 0.97³ = 0.912673 ≈ 0.9127.
    • (d) 1 − 0.97³ = 1 − 0.9127 = 0.0873.