Math 2030 A Homework 6. Answers and Solutions. Section 2.1: 4. We need to
calculate P(two sixes in the first five rolls | three sixes in eight rolls). This eval-.
Math 2030 A Homework 6 Answers and Solutions
Section 2.1: 4. We need to calculate P (two sixes in the first five rolls | three sixes in eight rolls). This evaluates to C5,2 · C3,1 C5,2 (1/6)2 (5/6)2 · C3,1 (1/6)(5/6)2 = .053714 . = 3 5 C8,3 (1/6) (5/6) C8,3 7. The probability that you win a single play is 15/36. The probability that you win 4 or 5 times in 5 plays is C5,4 (15/36)4 (21/36) + C5,5 (15/36)5 = 0.100468 .
Section 2.2: 4. Let Y be the number helped. Then Y ∼ Bin (300, 1/3). Approximate Y by X ∼ N (100, 200/3). Then P (Y > 120) is approximated by P (X > 120.5) ! 120.5 − 100 =P Z> p = 1 − Φ(2.51) = .006 . 200/3 8. Approximate this probability by P (99.5 ≤ X ≤ 100.5) where X ∼ N (100, 500/6). This is ! 99.5 − 100 100.5 − 100 p = P (−.055 ≤ Z ≤ .055) P ≤Z≤ p 500/6 500/6 = Φ(.055) − (1 − Φ(.055)) = .0438 . 9. (a) The number who show up follows Bin(324, 0.9) . Approximate with X ∼ N ((324)(0.9), (324)(0.9)((0.1)) = N (291.6, 29.6). 300.5 − 291.6 P (X > 300.5) = 1 − P Z ≤ = 1 − Φ(1.65) = 0.0495 . 5.4 (c) The number of pairs who show up follows Bin(162, 0.9). Approximate with X ∼ N ((162)(0.9), (162)(0.9)(0.1)) = N (145.8, 14.58) . 150.5 − 145.8 P (X > 150.5) = 1 − P Z ≤ = 1 − Φ(1.23) = .1093 . 3.82
