APPENDIX A
PROBABILITY CONCEPTS AND APPLICATIONS
SOLUTIONS TO PROBLEMS
A-15.
(a) P(H) = 1/2 = 0.5.
(b) P(T | H) = P(T) = 0.5.
A-16. The distribution of chips is as follows:
(a) The probability of drawing a white chip on the first draw is P(W) = 2/20 = 0.10.
A-17.
(a) The probability of getting a 4-inch nail is = 0.26.
A-18.
(a) The probability that an employee will have a cold next year is = 0.20.
A-19. The probability that the team wins tonight and draws a large crowd at tomorrow’s game is a joint
probability of dependent events. We know that the probability that the team wins tonight, P(W), is 0.60,
A-20. The second draw is not independent of the first because the probabilities of each outcome depend
on the rank (sophomore or junior) of the first student’s name drawn. Let J1 = junior on first draw, J2 =
junior on second draw, S1 = sophomore on first draw, and S2 = sophomore on second draw.
A-21. Without any additional information, we assume that there is an equally likely probability that the
soldier wandered into either oasis, so P(Abu Ilan) = 0.50 and P(El Kamin) = 0.50. Since the oasis of Abu
Ilan has 20 Bedouins and 20 Farimas (a total population of 40 tribesmen), the probability of finding a
A-22. P(Abu Ilan) is 0.50 and P(El Kamin) is 0.50.
P(2 Bedouins | Abu Ilan) = (0.50)(0.50) = 0.25.
A-23. P(Adjusted) = 0.8. P(Not adjusted) = 0.2. P(Pass | Adjusted) = 0.9. P(Pass | Not adjusted) = 0.2.
P(Adjusted and Pass) = P(Pass | Adjusted) P(Adjusted) = (0.9)(0.8) = 0.72.
A-24.
0.32 = 0.44.
(d) Probability = 1 – Winning every game = 1 – Answer to part (a) = 1 – 0.08 = 0.92.
A-25.
A-26.
(a) P(D or S) = P(D) + P(S) – P(D) P(S) = 0.90 + 0.95 – (0.90)(0.95) = 0.995.
A-27. In the sample of 1,000 people, 650 people were from Laketown and 350 from River City. Thirteen
A-28. P(F | 3) = [P(3 | F)P(F)] / [P(3 | F) P(F) + P(B | F’)P(F’)] = [(0.166)(0.5)]/[(0.166)(0.5)+(0.6)(0.5)]
A-29. Items (a) and (c) are valid probability distributions because the probability values for each event are
A-31.
X
P(X)
X P(X)
X – E(X)
[X – E(X)]2
[X – E(X)]2 P(X)
1
0.05
0.05
-4.45
19.80
0.990
2
0.05
0.10
-3.45
11.90
0.595
3
0.10
0.30
-2.45
6.00
0.600
4
0.10
0.40
-1.45
0.210
5
0.15
0.75
-0.45
0.20
0.030
6
0.15
0.90
0.045
7
0.25
1.75
1.55
2.40
0.601
8
0.15
1.20
0.975
5.45
4.046
A-32. Z = (X – ) / = (280-250)/25 = 1.20. The area under the curve corresponding to a Z of 1.20 is
A-33. Z = (X – ) / = (265-250)/25 = 0.60. The area under the curve to the left of Z = 0.60 is 0.7257.
Since we want to find the probability of selling more than 265 boats, we need the area to the right of Z =
0.5. Thus, the probability of selling fewer than 250 boats = 0.5.
A-34. For X = 0.65, Z = (0.65-0.55)/0.10 = 1. The area under the curve to the left of Z = 1 is 0.8413. For
A-35. P(X ≥ 0.65) = P(Z ≥ 1) = 0.1587.
A-36. μ = 450. σ = 25. P(X ≥ 475) = P(Z ≥ (475-450)/25) = P(Z ≥ 1) = 0.1587.
A-37. μ = 4,700. σ = 500.
(a) P(X ≥ 5,500) = P(Z ≥ 1.6) = 0.0548.
A-39. Ninety percent of the time, sales have been between 460,000 and 454,000 pencils. This means that
A-40. λ = 5 per day. e-5 = 0.0067.
(a) P(0) = (X e– / X!) = (50)(0.0067) / 0! = (1)(0.0067) / 1 = 0.0067.
A-42. μ = 3 per hour