 P(L′) = P(S)
 P(M $\cup $ S)
 P(F $\cap $ L)
 P(M$$L)
 P(L$$M)
 P(S$$F)
 P(F$$L)
 P(F $\cup $ L)
 P(M $\cap $ S)
 P(F)
In words, what is S  Z? : S  Z means, given that the customer has ordered pizza, the person also orders a salad.
In words, what is $ZANDS$? : $ZANDS$ represents the event that a customer orders a pizza and salad.
Are Z and S mutually exclusive events? Show why or why not. No, because P(S and Z) does not equal 0.
P(musician is a male $\cap $ had private instruction) = $\frac{15}{130}$ = $\frac{3}{26}$ = 0.12
The events are not mutually exclusive. It is possible to be a female musician who learned music in school.
To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is $\frac{\mathrm{4,715}}{\mathrm{100,450}}$.
To pick one person from the study who is Japanese American given that person smokes 2130 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 2130 cigarettes per day. The probability is $\frac{4715}{\mathrm{15,273}}$.
 P(GG) = $\left(\frac{5}{8}\right)\left(\frac{5}{8}\right)$ = $\frac{25}{64}$
 P(at least one green) = P(GG) + P(GY) + P(YG) = $\frac{25}{64}$ + $\frac{15}{64}$ + $\frac{15}{64}$ = $\frac{55}{64}$
 P(G$$G) = $\frac{5}{8}$
 Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.

<20 20–64 >64 Totals Female 0.0244 0.3954 0.0661 0.486 Male 0.0259 0.4186 0.0695 0.514 Totals 0.0503 0.8140 0.1356 1  P(F) = 0.486
 P(>64 $$ F) = 0.1361
 P(>64 and F) = P(F) P(>64F) = (0.486)(0.1361) = 0.0661
 P(>64 $$ F) is the percentage of female drivers who are 65 or older and P(>64 $\cap $ F) is the percentage of drivers who are female and 65 or older.
 P(>64) = P(>64 $\cap $ F) + P(>64 $\cap $ M) = 0.1356
 No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 $\cap $ F) = 0.0661.

Car, truck or van Walk Public transportation Other Totals Alone 0.7318 Not alone 0.1332 Totals 0.8650 0.0390 0.0530 0.0430 1  If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = 0.7318 + 0.0390 = 0.7708.
 Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771
 (0.1332)(1,000) = 133
 You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%
 A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.
 The Forum Research surveyed 1,046 Torontonians.
 58%
 42% of 1,046 = 439 (rounding to the nearest integer)
 0.57
 0.60.
 P(Betting on two line that touch each other on the table) = $\frac{6}{38}$
 P(Betting on three numbers in a line) = $\frac{3}{38}$
 P(Betting on one number) = $\frac{1}{38}$
 P(Betting on four number that touch each other to form a square) = $\frac{4}{38}$
 P(Betting on two number that touch each other on the table ) = $\frac{2}{38}$
 P(Betting on 000123) = $\frac{5}{38}$
 P(Betting on 012; or 0002; or 0023) = $\frac{3}{38}$
 {G1, G2, G3, G4, G5, Y1, Y2, Y3}
 $\frac{5}{8}\text{}$
 $\frac{2}{3}\text{}$
 $\frac{2}{8}\text{}$
 $\frac{6}{8}\text{}$
 No, because P(G $\cap $ E) does not equal 0.
NOTE
The coin toss is independent of the card picked first.
 {(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)}
 P(A) = P(blue)P(head) = $\left(\frac{3}{10}\right)$$\left(\frac{1}{2}\right)$ = $\frac{3}{20}$
 Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P(A $\cap $ B) = 0
 No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A $\cap $ C) = P(A) = $\frac{3}{20}$
 S = {(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
 $\frac{4}{8}$
 Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P(A $\cap $ B) = 0.
 If Y and Z are independent, then P(Y $\cap $ Z) = P(Y)P(Z), so P(Y $\cup $ Z) = P(Y) + P(Z)  P(Y)P(Z).
 0.5
 P(R) = 0.44
 P(R$$E) = 0.56
 P(R$$O) = 0.31
 No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P(R$$E) ≠ P(R).
 No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P(R$$E) > P(R).
P(type O OR Rh) = P(type O) + P(Rh) – P(type O AND Rh)
0.38 = 0.45 + 0.07 – P(type O AND Rh);
Solve to find P(type O AND Rh) = 0.14.
14% of people have type O, Rh blood.P(NOT(type O AND Rh)) = 1 – P(type O AND Rh) = 1 – 0.14 = 0.86
86% of people do not have type O, Rh blood.
 Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts.
 P(C $\cup $ N) = P(C) + P(N)  P(C $\cap $ N) = 0.36 + 0.12  0.08 = 0.40
 P(NEITHER chocolate NOR nuts) = 1  P(C $\cup $ N) = 1  0.40 = 0.60

Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 156 TOTALS 2,792 5,279 9,354 3,656 21,081 
Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 278 517 721 156 1672 TOTALS 2,792 5,279 9,354 3,656 21,081  $\frac{\text{8,395}}{\text{21,081}}\approx 0.3982$
 $\frac{\text{1,061}}{\text{21,081}}\approx 0.0503$
 $\frac{\text{1,885}}{\text{21,081}}\approx 0.0894$
 $\frac{\text{9,219}}{\text{21,081}}\approx 0.4373$
 $\frac{\text{1,595}}{\text{3,656}}\approx 0.4363$
 $\frac{26}{106}$
 $\frac{33}{106}$
 $\frac{21}{106}$
 $\left(\frac{26}{106}\right)$ + $\left(\frac{33}{106}\right)$  $\left(\frac{21}{106}\right)$ = $\left(\frac{38}{106}\right)$
 $\frac{21}{33}$
 P(C) = 0.4567
 not enough information
 not enough information
 No, because over half (0.51) of men have at least one false positive text
 $(J\cup K)=P(J)+P\left(K\right)P(J\cap K);$ $0.45=0.18+0.37P(J\cap K);$ solve to find $P(J\cap K)=0.10$
 $P\left(\text{NOT}\right(J\cap K\left)\right)=1P(J\cap K)$ $=10.10=0.90$
 $P\left(\text{NOT}\right(J\cup K\left)\right)=1P(J\cup K)$ $=10.45=0.55$