MAT-130: Module 6 Exam

$ Let: \\[3ex] head = H \\[3ex] tail = T \\[3ex] S = \{75H, 25T\} \\[3ex] n(S) = 100 \\[3ex] n(H) = 75 \\[3ex] P(H) = \dfrac{n(H)}{n(S)} \\[5ex] P(H) = \dfrac{75}{100} \\[5ex] P(H) = 0.75 \\[3ex] $ The probability that the next flip results in a head is $0.75$

$ Let: \\[3ex] Advanced\:\:Degree = A \\[3ex] P(A) = 13.1\% = \dfrac{13.1}{100} \\[5ex] P(A) = 0.131 \\[3ex] $ The probability that a randomly selected U.S adult has an advanced degree is $0.131$

(a.)

Type of Larceny Theft	Number of Offenses	Probability
Pocket picking	$6$	$\approx 0.010$
Purse snatching	$7$	$\approx 0.012$
Shoplifting	$120$	$\approx 0.207$
From motor vehicles	$201$	$\approx 0.347$
Motor vehicle accessories	$85$	$\approx 0.147$
Bicycles	$35$	$\approx 0.060$
From buildings	$117$	$\approx 0.202$
From coin-operated machines	$8$	$\approx 0.014$
	$\Sigma Offenses = 579$	$\Sigma Probability \approx 1$

$ (b.) \\[3ex] P(Pocket\:\:picking) = 0.010 \\[3ex] 0.010 \lt 0.05 \\[3ex] $ Pocket-picking larcenies are unusual because the probability of pocket picking larceny is less than $5\%$ ... Usual and Unusual Events

$ (c.) \\[3ex] P(From\:\:buildings) = 0.202 \\[3ex] 0.202 \gt 0.05 \\[3ex] $ Larcenies from buildings are not unusual because the probability of a larceny from a building is greater than $5\%$ ... Usual and Unusual Events

$ AND \implies \cap \\[3ex] (a.) \\[3ex] A \cap B = \{\} \\[3ex] A \cap B = \phi \\[3ex] n(A \cap B) = 0 \\[3ex] P(A \cap B) = 0 \\[3ex] $ Events $A$ and $B$ are disjoint events because they do not have any common outcome.
Events $A$ and $B$ are also mutually exclusive events because they cannot occur at the same time.

$ OR \implies \cup \\[3ex] P(C) = 0.25 \\[3ex] P(D) = 0.48 \\[3ex] P(C \cup D) = P(C) + P(D) - P(C \cap D)...Addition\:\:Rule \\[3ex] P(C \cap D) = 0 ...Mutually\:\:Exclusive\:\:Events \\[3ex] \rightarrow P(C \cup D) = 0.25 + 0.48 \\[3ex] P(C \cup D) = 0.73 \\[3ex] $ The probability that events $C$ OR $D$ will occur is $0.73$

$ OR \implies \cup \\[3ex] Let: \\[3ex] green = G \\[3ex] brown = B \\[3ex] red = R \\[3ex] S = \{7G, 3G, 8R\} \\[3ex] n(S) = 7 + 3 + 8 \\[3ex] n(S) = 18 \\[3ex] n(G) = 7 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} \\[5ex] P(G) = \dfrac{7}{18} \\[5ex] n(B) = 3 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} \\[5ex] P(B) = \dfrac{3}{18} \\[5ex] P(G \cup B) = P(G) + P(B) - P(G \cap B) ...Addition\:\:Rule \\[3ex] \underline{Single\:\:pick} \\[3ex] P(G \cap B) = 0 ...Mutually\:\:Exclusive\:\:Events \\[3ex] \rightarrow P(G \cup B) = \dfrac{7}{18} + \dfrac{3}{18} - 0 \\[5ex] P(G \cup B) = \dfrac{7 + 3}{18} \\[5ex] P(G \cup B) = \dfrac{10}{18} \\[5ex] P(G \cup B) = \dfrac{5}{9} \\[5ex] P(G \cup B) = 0.555555555556 \\[3ex] P(G \cup B) \approx 0.556 \\[3ex] $ The probability that Cornelius will select a green OR brown marble in a single pick of a marble from the bag is $0.556$

$ Let: Great\:\:Dane\:\:lives\:\:to\:\:be\:\:7\:\:years = L \\[3ex] \implies Great\:\:Dane\:\:does\:\:not\:\:live\:\:to\:\:be\:\:7\:\:years = L' \\[3ex] P(L) = 0.98156 \\[3ex] P(L) + P(L') = 1 ...Complementary\:\:Events \\[3ex] P(L') = 1 - P(L) \\[3ex] P(L') = 1 - 0.98156 \\[3ex] P(L') = 0.01844 \\[3ex] (a.) \\[3ex] P(LL) = (0.98156)(0.98156) ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(LL) = 0.9634600336 \\[3ex] P(LL) \approx 0.96346 \\[3ex] (b.) \\[3ex] P(LLLLLLLL) = (0.98156)^8 ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(LLLLLLLL) = 0.861657784 \\[3ex] P(LLLLLLLL) \approx 0.86166 \\[3ex] $ (c.)
We can solve this in two ways - Multiplication and Addition Rules and the Complementary Rule

First Method - Multiplication and Addition Rules
At least $1$ implies $1$ OR More
This means that the first dog will not live to be $7$ years old AND the seven other dogs (second through the eighth) will live OR the second dog will not live AND the first dog, the third dog, the fourth dog through the eighth dog will live, OR the first and second dog will not live AND the third dog through the eighth dog will live OR ....and it keeps going on...
This is a Herculean task.
So, we shall solve it using the second method.

Second Method - Complementary Rule
Event: at least one dog will not live ($1$ or more dogs will not live)
Complementary Event: All dogs will live (all eighth dogs will live)

$ P(LLLLLLLL) = 0.861657784 \\[3ex] (c.) \\[3ex] P(at\:\:least\:\:L') = 1 - P(LLLLLLLL) \\[3ex] P(at\:\:least\:\:L') = 1 - 0.861657784 \\[3ex] P(at\:\:least\:\:L') = 1 - 0.861657784 \\[3ex] P(at\:\:least\:\:L') = 0.138342216 \\[3ex] P(at\:\:least\:\:L') \approx 0.13834 \\[3ex] (d.) \\[3ex] 0.13834 \gt 0.05 \\[3ex] $ It is not unusual if at least one of eight randomly selected Great Danes did not live to be $7$ years old because the probability that at least one of eight randomly selected Great Danes will not live to be $7$ years is greater than $5\%$ ... Usual and Unusual Events

$ Let:\:\:Pasteurized\:\:cheese = P \\[3ex] Let\:\:Raw-milk\:\:cheese = R \\[3ex] \implies NOT\:\:Pasteurized\:\:cheese = P' \\[3ex] P(P') = P(R) ...Based\:\:on\:\:the\:\:Question \\[3ex] P(P) = 88\% = \dfrac{88}{100} = 0.88 \\[5ex] P(P) + P(P') = 1 ...Complementary\:\:Events \\[3ex] P(P') = 1 - P(P) \\[3ex] P(P') = 1 - 0.88 \\[3ex] P(P') = 0.12 \\[3ex] P(R) = 0.12 \\[3ex] (a.) \\[3ex] P(PP) = (0.88)(0.88) ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(PP) = 0.7744 \\[3ex] (b.) \\[3ex] P(PPPP) = (0.88)(0.88)(0.88)(0.88) = 0.88^4 ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(PPPP) = 0.59969536 \\[3ex] P(PPPP) \approx 0.5997 \\[3ex] $ (c.)
We can solve this in two ways - Multiplication and Addition Rules and the Complementary Rule

First Method - Multiplication and Addition Rules
At least $1$ implies $1$ OR More
At least one of four cheeses is raw-milk means that $1$ or more is not pasteurized

This means that the first is raw-milk AND the second, third, and fourth are NOT: $(RPPP)$, OR

the first and second are raw-milk AND the third and fourth are NOT: $(RRPP)$ OR

the first, second, and third milk are raw-milk AND the fourth is NOT: $(RRRP)$ OR

all four are raw-milk $(RRRR)$ OR

the first is NOT raw-milk AND the second, third and fourth are: $(PRRR)$ OR

the first and second are NOT raw-milk AND the third and fourth are: $(PPRR)$ OR

....and it keeps going on and on and on...
This is a Herculean task.
So, we shall solve it using the second method.

Second Method - Complementary Rule
Event: at least one cheese is raw-milk ($1$ or more milk containers are raw-milk)
Complementary Event: All cheeses are pasteurized (all cheeses are NOT raw-milk)

$ P(PPPP) = 0.59969536 \\[3ex] (c.) \\[3ex] P(at\:\:least\:\:R) = 1 - P(PPPP) \\[3ex] P(at\:\:least\:\:R) = 1 - 0.59969536 \\[3ex] P(at\:\:least\:\:R) = 0.40030464 \\[3ex] P(at\:\:least\:\:R) \approx 0.4003 \\[3ex] (d.) \\[3ex] 0.4003 \gt 0.05 \\[3ex] $ It is not unusual if at least one of four randomly selected cheeses is raw-milk because the probability that at least one of four randomly selected cheeses is raw-milk is greater than $5\%$ ... Usual and Unusual Events

$ Let: \\[3ex] Default\:\:mortgage = D \\[3ex] No\:\:default\:\:mortgage = D' \\[3ex] P(D) = 0.18 \\[3ex] (a.) \\[3ex] P(D) + P(D') = 1 ...Complementary\:\:Rule \\[3ex] P(D') = 1 - P(D) \\[3ex] P(D') = 1 - 0.18 \\[3ex] P(D') = 0.82 \\[3ex] (b.) \\[3ex] P(D'D'D'D'D') = 0.82 * 0.82 * 0.82 * 0.82 * 0.82 ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(D'D'D'D'D') = 0.82^5 \\[4ex] P(D'D'D'D'D') = 0.3707398432 \\[3ex] P(D'D'D'D'D') \approx 0.3707 \\[3ex] $ (c.)
We can solve this in two ways - Multiplication and Addition Rules and the Complementary Rule

First Method - Multiplication and Addition Rules
At least $1$ implies $1$ OR More
At least one of the five mortgages defaults means that $1$ or more mortgages defaults

This means that the first mortgage defaults AND the second, third, fourth, and fifth does NOT: $(DD'D'D'D')$, OR

the first and second mortgages defaults AND the third, fourth, and fifth does NOT: $(DDD'D'D')$ OR

the first, second, and third mortgages defaults AND the fourth and fifth does NOT: $(DDDD'D')$ OR

the first, second, third,a nd fourth mortgages default AND the fifth does NOT: $(DDDDD')$ OR

all five mortgages defaults $(D'D'D'D'D')$ OR

the first does NOT default AND the second, third, fourth and fifth defaults: $(D'DDDD)$ OR

the first and second does not default AND the third, fourth, and fifth defaults are: $(D'D'DDD)$ OR

....and it keeps going on and on and on...it is much work

So, we shall solve it using the second method.

Second Method - Complementary Rule
Event: at least one mortgage defaults ($1$ or more mortgages defaults)
Complementary Event: All mortgages do NOT default

$ P(D'D'D'D'D') = 0.3707398432 \\[3ex] (c.) \\[3ex] P(at\:\:least\:\:D) = 1 - P(D'D'D'D'D') \\[3ex] P(at\:\:least\:\:D) = 1 - 0.3707398432 \\[3ex] P(at\:\:least\:\:R) = 0.6292601568 \\[3ex] P(at\:\:least\:\:R) \approx 0.6293 \\[3ex] $ The probability that at least one of the mortgages defaults is $0.6293$

$ Let: \\[3ex] Cloudy\:\:day = C \\[3ex] Foggy\:\:day = F \\[3ex] P(C) = 72\% = \dfrac{72}{100} = 0.72 \\[5ex] P(C \cap F) = 58\% = \dfrac{58}{100} = 0.58 \\[5ex] P(F | C) = ? \\[3ex] P(F | C) = \dfrac{P(F \cap C)}{P(C)} ...Conditional\:\:Probability \\[5ex] P(F | C) = \dfrac{0.58}{0.72} \\[5ex] P(F | C) = 0.805555556 \\[3ex] P(F | C) \approx 0.806 \\[3ex] $ The probability that a randomly selected day in October will be foggy if it is cloudy is $0.806$

$ Let: \\[3ex] Driver\:\:aged\:\:55-69 = D \\[3ex] n(D) = 5772 + 2469 = 8241 \\[3ex] Male = M \\[3ex] n(M) = 155 + 5548 + 11712 + 11498 + 5772 + 3324 = 38009 \\[3ex] n(M \cap D) = 5772 \\[3ex] (a.) \\[3ex] P(D | M) = \dfrac{n(D \cap M)}{n(M)} \\[5ex] P(D | M) = \dfrac{5772}{38009} \\[5ex] P(D | M) = 0.151858770 \\[3ex] P(D | M) \approx 0.152 \\[3ex] (b.) \\[3ex] P(M | D) = \dfrac{n(M \cap D)}{n(D)} \\[5ex] P(M | D) = \dfrac{5772}{8241} \\[5ex] P(M | D) = 0.700400437 \\[3ex] P(M | D) \approx 0.700 \\[3ex] (c.) \\[3ex] 0.700 \gt 0.5 \\[3ex] $ The victim of a fatal accident aged $55$ to $69$ years old is more likely to be a male driver because the probability that a randomly selected driver fatality who was $55$ to $69$ years old was a male is greater than $50\%$

$ n(S) = 52 \\[3ex] Let\:\:spade = P \\[3ex] n(P) = 13 \\[3ex] P(P) = \dfrac{n(P)}{n(S)} \\[5ex] P(P) = \dfrac{13}{52} \\[5ex] (a.) \\[3ex] Without\:\:Replacement \implies Dependent\:\:Events \\[3ex] P(PP) = \dfrac{13}{52} * \dfrac{12}{51} ...Multiplication\:\:Rule \\[5ex] P(PP) = \dfrac{1}{17} \\[5ex] P(PP) = 0.058823529 \\[3ex] P(PP) \approx 0.059 \\[3ex] (b.) \\[3ex] With\:\:Replacement \implies Independent\:\:Events \\[3ex] P(PP) = \dfrac{13}{52} * \dfrac{13}{51} ...Multiplication\:\:Rule \\[5ex] P(PP) = \dfrac{1}{4} * \dfrac{1}{4} \\[5ex] P(PP) = \dfrac{1}{16} \\[5ex] P(PP) = 0.0625 \\[3ex] P(PP) \approx 0.063 $

This is a case of Combination because the samples can be obtained without regard to order of selection

$ _nC_r = \dfrac{n!}{(n - r)!r!} \\[5ex] For\:\: _{52}C_4 \\[3ex] n = 52 \\[3ex] r = 4 \\[3ex] _{52}C_4 = \dfrac{52!}{(52 - 4)! 4!} \\[5ex] _{52}C_4 = \dfrac{52!}{48! 4!} \\[5ex] _{52}C_4 = \dfrac{52!}{48! * 4!} \\[5ex] _{52}C_4 = \dfrac{52 * 51 * 50 * 49 * 48!}{48! * 4 * 3 * 2 * 1} \\[5ex] _{52}C_4 = 13 * 17 * 25 * 49 \\[3ex] _{52}C_4 = 270,725 \\[3ex] $ $270,725$ different simple random samples of size $4$ can be obtained from a population of size $52$

This is a case of Permutation because the five balls must be matched in the correct order (the order of arrangement is important)

$ _nP_r = \dfrac{n!}{(n - r)!} \\[5ex] For\:\: _{24}C_5 \\[3ex] n = 24 \\[3ex] r = 5 \\[3ex] _{24}C_5 = \dfrac{24!}{(24 - 5)!} \\[5ex] _{24}C_5 = \dfrac{24!}{19!} \\[5ex] _{24}C_5 = \dfrac{24 * 23 * 22 * 21 * 20 * 19!}{19!} \\[5ex] _{24}C_5 = 24 * 23 * 22 * 21 * 20 \\[3ex] _{24}C_5 = 5,100,480 \\[3ex] $ The number of possible outcomes for the lottery is $5,100,480$ outcomes

$ AND \implies \cap \\[3ex] P(B | A) = \dfrac{P(B \cap A)}{P(A)} \\[5ex] P(B \cap A) = P(A \cap B) \\[3ex] P(B | A) = \dfrac{P(A \cap B)}{P(A)} \\[5ex] \dfrac{P(A \cap B)}{P(A)} = P(B | A) \\[5ex] P(A \cap B) = P(B | A) * P(A) \\[3ex] P(A \cap B) = 0.5 * 0.4 \\[3ex] P(A \cap B) = 0.2 $