Problem 1

a. S = {(F,F,F,F), (F,F,F,V), (F,F,V,F), (F,V,F,F), (V,F,F,F), (F,F,V,V,), (F,V,F,V), (F,V,V,F), (V,F,F,V), (V,F,V,F,), (V,V,F,F,), (F,V,V,V), (V,F,V,V), (V,V,F,V), (V,V,V,F), (V,V,V,V)}

b. A = {(F,V,V,V), (V,F,V,V), (V,V,F,V), (V,V,V,F), (V,V,V,V)}

c. B = {(F,F,F,F), (V,V,V,V)}

d. C = A U B,
C = {(F,V,V,V), (V,F,V,V), (V,V,F,V), (V,V,V,F), (V,V,V,V), (F,F,F,F)}

e. D = A Π B,
D = {(V,V,V,V)}

Problem 2

a. P(brown) = 0.15

b. P(red or green) = P(red) + P(green) = 0.10 + 0.15 = 0.25

c. P(not blue) = 1 – P(blue) = 1- 0.25 = 0.75

d. P(red and brown) = 0

Problem 3

a.

library(ggplot2)


y <- 0:10
p <- c(0.100, 0.130, 0.250, 0.160, 0.095, 0.075, 0.063, 0.047, 0.041, 0.024, 0.015)

df <- data.frame(y,p)

ggplot(df, aes(x=factor(y), y=p)) + 
  geom_bar(stat = "identity", fill="skyblue", color="black") +
  labs(title="Probability Distribution of Y", x= "Number of Cars Failing (y)",
       y= "Probability P(Y=y)")

b. P(Y≤2) = P(Y=0) + P(Y=1) + P(Y=2) = 0.1 + 0.13 + 0.25 = 0.48

c. P(Y>7) = P(Y=8) + P(Y=9) + P(Y=10) = 0.041 + 0.024 + 0.015 = 0.080

d. P(2<Y≤7) = P(Y=3) + P(Y=4) + P(Y=5) + P(Y=6) + P(Y=7) = 0.16 + 0.095 + 0.075 + 0.063 + 0.047 = 0.44

Problem 4

a. P(A and T1) = 0.017

b. P(T2| W) = P (W and T2) / P (W) = 0.322/0.802 = 0.401

P(not T2 | W) = 1- 0.401 = .599

c. P(T1 or T2 | A) = (0.012 + 0.01)/ 0.042 = 0.524

d. P(neither T2 nor T4) = 1 – P (T2 or T4)

P(T2)= 0.322 + 0.029 + 0.012 + 0.008 = .371

P(T4)= 0.032 + 0.005 + 0.003 + 0.001 = 0.041

P (T2 or T4) = 0.371 + 0.041 = 0.412

P(neither T2 nor T4) = 1 – 0.412 = 0.588

e. The event that the donor has blood type O, given the donor is white.

f. P( T1 | W) = P(T1 and W) / P(W) = 0.36 / 0.802 = 0.449

g. The events W and T1 are not independent.

P(T1 | W) ≠ P (T1)

0.449 ≠ P (T1) = 0.36 + 0.07 + 0.017 + 0.015 = 0.462

0.449 ≠ 0.462

h. W and T1 are not mutually exclusive; they can occur at the same time. A donor can be white and have blood type O. P(W and T1) is a given value in the chart, 0.360.

Problem 5

a. Continuous

b. Discrete

c. Continuous

d. Continuous

e. Discrete

f. Discrete

Problem 6

a. P (Y=15) = (0.20)15

dbinom(15, 15, 0.2) = 3.2768e-11

dbinom(15, 15, 0.2)
## [1] 3.2768e-11

b. P (Y = 6) = (15!/ (6! * 9!)) (.2)6 (1 – 0.2)9

dbinom(6, 15, 0.2) = 0.429

dbinom(6, 15, 0.2)
## [1] 0.04299262
P (Y ≥ 6) = ∑_(k=6)^15▒〖(15!/(6!*(15-k)!)) (0.2)^k (1-0.2)^(15-k) 〗

c. 1 - pbinom (5, 15, 0.2) = .061

1 - pbinom(5, 15, 0.2)
## [1] 0.06105143

d. P (Y = 0) = (0.8)15

dbinom(0, 15, 0.2) = 0.352

dbinom(0,15,0.2)
## [1] 0.03518437