Here $E=\{4\}$ and $F=\{2,4,6\}$. Since $E\cap F=\{4\}\ne\varnothing$, the events can occur together and are not mutually exclusive.
No, E and F are not mutually exclusive.
The sample space is $S=\{1,2,3,4,5,6\}$. Write each event as a subset of $S$ and then apply union, intersection, difference and complement operations directly.
$A=\{1,2,3,4,5,6\}$, $B=\varnothing$, $C=\{3,6\}$, $D=\{1,2,3\}$, $E=\{6\}$, $F=\{3,4,5,6\}$; $A\cup B=A$, $A\cap B=\varnothing$, $B\cup C=\{3,6\}$, $E\cap F=\{6\}$, $D\cap E=\varnothing$, $A-C=\{1,2,4,5\}$, $D-E=\{1,2,3\}$, $E\cap F'=\varnothing$, $F'=\{1,2\}$.
List ordered pairs from two dice. Event $C$ consists of sums 9 and 12. Since $C\subset A$, $A$ and $C$ are not mutually exclusive. No outcome in $B$ has sum greater than 8 or sum 9 or 12, so $A\cap B=\varnothing$ and $B\cap C=\varnothing$.
$A=\{(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)\}$; $B=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2)\}$; $C=\{(3,6),(4,5),(5,4),(6,3),(6,6)\}$. The mutually exclusive pairs are $(A,B)$ and $(B,C)$.
Here $A=\{HHH\}$, $B=\{HHT,HTH,THH\}$, $C=\{TTT\}$ and $D=\{HHH,HHT,HTH,HTT\}$. Simple events contain one sample point; compound events contain more than one. Mutually exclusive pairs have empty intersection.
(i) Mutually exclusive pairs: $A$ and $B$, $A$ and $C$, $B$ and $C$, $C$ and $D$. (ii) Simple events: $A$ and $C$. (iii) Compound events: $B$ and $D$.
For three coin tosses, $S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$. The listed examples are checked by intersections: mutually exclusive events have no common outcomes, and exhaustive events have union equal to $S$.
One possible answer: (i) $A=$ no head and $B=$ exactly one head. (ii) no head, exactly one head, at least two heads. (iii) at least one head and at least one tail. (iv) no head and three heads. (v) no head, exactly one head, exactly two heads.
Since $A$ and $B$ are complementary events, $A'=B$ and $B'=A$. Event $C'$ means the sum is greater than 5. Substitute these descriptions into each requested expression.
(i) odd number on the first die; (ii) even number on the first die; (iii) all outcomes; (iv) impossible event; (v) even first die and sum greater than 5; (vi) odd first die or sum at most 5; (vii) odd first die and sum at most 5; (viii) even first die and sum greater than 5.
$A$ is even first die and $B$ is odd first die, so they are complements. Thus (i), (ii) and (iii) are true. For (iv), $A\cap C$ is not empty, for example $(2,1)$. For (v), $B'=A$, so $A\cap B'=A\ne\varnothing$. For (vi), $A'=B$ and $B'=A$ overlap with $C$, so the three events are not mutually exclusive, though their union covers the sample space.
(i) True; (ii) True; (iii) True; (iv) False; (v) False; (vi) False.
A valid probability assignment must have all probabilities non-negative and total probability 1. In (c), the sum is $0.1+0.2+\cdots+0.7=2.8>1$. In (d), some probabilities are negative. In (e), the sum is $\dfrac{1+2+3+4+5+6+15}{14}=\dfrac{36}{14}>1$. Assignments (a) and (b) are valid.
Assignments (c), (d) and (e) cannot be valid.
The sample space is $\{HH,HT,TH,TT\}$. Outcomes with at least one tail are $HT,TH,TT$, so the probability is $\dfrac{3}{4}$.
$\dfrac34$.
For a die, $S=\{1,2,3,4,5,6\}$. Prime outcomes are $\{2,3,5\}$. Outcomes $\ge3$ are $\{3,4,5,6\}$. Outcomes $\le1$ are $\{1\}$. No outcome is more than 6. Outcomes less than 6 are $\{1,2,3,4,5\}$.
(i) $\dfrac12$; (ii) $\dfrac23$; (iii) $\dfrac16$; (iv) $0$; (v) $\dfrac56$.
There are 52 cards. The ace of spades is one card, so probability $=1/52$. There are 4 aces, so probability of an ace $=4/52=1/13$. There are 26 black cards, so probability $=26/52=1/2$.
(a) $52$; (b) $\dfrac1{52}$; (c)(i) $\dfrac1{13}$; (c)(ii) $\dfrac12$.
There are $2\times6=12$ equally likely outcomes. Sum 3 occurs only when the coin shows 1 and the die shows 2. Sum 12 occurs only when the coin shows 6 and the die shows 6. Each probability is $\dfrac1{12}$.
(i) $\dfrac1{12}$; (ii) $\dfrac1{12}$.
There are $4+6=10$ council members, of whom 6 are women. Probability $=\dfrac6{10}=\dfrac35$.
$\dfrac35$.
If $h$ heads occur in four tosses, tails are $4-h$ and the amount is $h(1)-(4-h)(1.50)=2.5h-6$. For $h=0,1,2,3,4$, this gives Rs $-6,-3.50,-1,1.50,4$. The probabilities are $\binom4h/16$, i.e. $1/16,4/16,6/16,4/16,1/16$.
The possible amounts are Rs $-6,-3.50,-1,1.50,4$ with probabilities $\dfrac1{16},\dfrac4{16},\dfrac6{16},\dfrac4{16},\dfrac1{16}$ respectively.
There are 8 equally likely outcomes. Counts are: 3 heads: 1; exactly 2 heads: 3; at least 2 heads: 4; at most 2 heads: 7; no head: 1; 3 tails: 1; exactly 2 tails: 3; no tail: 1; at most 2 tails: 7. Divide each count by 8.
(i) $\dfrac18$; (ii) $\dfrac38$; (iii) $\dfrac12$; (iv) $\dfrac78$; (v) $\dfrac18$; (vi) $\dfrac18$; (vii) $\dfrac38$; (viii) $\dfrac18$; (ix) $\dfrac78$.
For any event $A$, $P(A')=1-P(A)$. Given $P(A)=\dfrac12$, $P(A')=1-\dfrac12=\dfrac12$.
$\dfrac12$.
The word ASSASSINATION has 13 letters. Vowels are A, A, I, A, I, O, i.e. 6 letters. Consonants are the remaining 7 letters. Therefore the probabilities are $6/13$ and $7/13$.
(i) $\dfrac6{13}$; (ii) $\dfrac7{13}$.
There are 60 marbles, so total ways to draw 5 are $\binom{60}{5}$. (i) All blue can be chosen in $\binom{20}{5}$ ways. (ii) At least one green is the complement of drawing all 5 from the 30 non-green marbles, so probability is $1-\dfrac{\binom{30}{5}}{\binom{60}{5}}$.
(i) $\dfrac{\binom{20}{5}}{\binom{60}{5}}$; (ii) $1-\dfrac{\binom{30}{5}}{\binom{60}{5}}$.
Total ways to draw 4 cards are $\binom{52}{4}$. Choose 3 diamonds from 13 and 1 spade from 13, giving $\binom{13}{3}\binom{13}{1}$ favourable draws. Hence the probability is $\dfrac{\binom{13}{3}\binom{13}{1}}{\binom{52}{4}}$.
$\dfrac{\binom{13}{3}\binom{13}{1}}{\binom{52}{4}}$.
The die has 6 faces: two labelled 1, three labelled 2, and one labelled 3. Thus $P(2)=3/6=1/2$, $P(1\text{ or }3)=(2+1)/6=1/2$, and $P(\text{not }3)=5/6$.
(i) $\dfrac12$; (ii) $\dfrac12$; (iii) $\dfrac56$.
There are 9990 non-prize tickets among 10000. For one ticket, probability of no prize is $9990/10000=999/1000$. For two tickets, both must come from the 9990 non-prize tickets, giving $\dfrac{\binom{9990}{2}}{\binom{10000}{2}}$. For 10 tickets, probability $=\dfrac{\binom{9990}{10}}{\binom{10000}{10}}$.
(a) $\dfrac{999}{1000}$; (b) $\dfrac{\binom{9990}{2}}{\binom{10000}{2}}$; (c) $\dfrac{\binom{9990}{10}}{\binom{10000}{10}}$.
If you are in the section of 40, your friend is in the same section with probability $39/99$; this case has probability $40/100$. If you are in the section of 60, your friend is in the same section with probability $59/99$; this case has probability $60/100$. Thus probability of same section is $\dfrac{40}{100}\dfrac{39}{99}+\dfrac{60}{100}\dfrac{59}{99}=\dfrac{17}{33}$. Probability of different sections is $1-\dfrac{17}{33}=\dfrac{16}{33}$.
(a) $\dfrac{17}{33}$; (b) $\dfrac{16}{33}$.
There are $3!=6$ ways to place the letters. The number of derangements of 3 letters is 2, so the probability that no letter is in the proper envelope is $2/6=1/3$. Therefore the probability that at least one letter is in its proper envelope is $1-1/3=2/3$.
$\dfrac23$.
$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.54+0.69-0.35=0.88$. Then $P(A'\cap B')=1-P(A\cup B)=0.12$. Also $P(A\cap B')=P(A)-P(A\cap B)=0.19$ and $P(B\cap A')=P(B)-P(A\cap B)=0.34$.
(i) $0.88$; (ii) $0.12$; (iii) $0.19$; (iv) $0.34$.
Males are Harish, Rohan and Salim. Persons over 35 are Sheetal and Salim. The union contains Harish, Rohan, Salim and Sheetal, i.e. 4 persons out of 5. Probability $=4/5$.
$\dfrac45$.
(i) With repetition, the first digit is 5 or 7, the middle two digits have 5 choices each, and the last digit must be 0 or 5. Favourable numbers $=2\cdot5\cdot5\cdot2=100$, total $=2\cdot5^3=250$, so probability $=2/5$. (ii) Without repetition, total numbers $=2\cdot{}^4P_3=48$. Favourable cases: if first digit is 5, last must be 0 and the middle two have $3P2=6$ choices; if first digit is 7, last can be 0 or 5, giving $6+6=12$ choices. Favourable $=18$, so probability $=18/48=3/8$.
(i) $\dfrac25$; (ii) $\dfrac38$.
The number of possible four-digit sequences with no repeats from 10 digits is ${}^{10}P_4=10\cdot9\cdot8\cdot7=5040$. Only one sequence opens the lock, so the probability is $\dfrac1{5040}$.
$\dfrac1{5040}$.