Probability

Let $E=\{4\}$ and $F=\{2,4,6\}$. Since $E\cap F=\{4\}\neq \varnothing$, the events are not mutually exclusive.

I can answer this only if the rest of the question is provided.

For two dice, the events are:

- A: sum $>8$.
- B: 2 occurs on either die.
- C: sum is at least 7 and a multiple of 3, so the sum is $9$ or $12$.

The mutually exclusive pairs are:
- A and C: if the sum is $9$ or $12$, it is not possible for the sum to be greater than $8$? Actually $9$ and $12$ are greater than $8$, so A and C are not mutually exclusive.
- A and B: they can happen together, for example $(2,6)$ gives sum $8$ not in A, but $(2,5)$ gives $7$ not in A; however $(2,6)$ no. Check a case with sum $>8$ and a 2, such as $(2,6)$ gives $8$, not enough; $(2,5)$ gives $7$; $(2,4)$ gives $6$. So A and B are mutually exclusive.
- B and C: can happen together, e.g. $(2,7)$ is impossible, but $(2,4)$ sum $6$ not in C. No common outcome with sum $9$ or $12$ containing a 2 among two dice? $(3,6)$ gives 9 but no 2. So B and C are mutually exclusive.

Thus, the mutually exclusive pairs are (A, B) and (B, C).

The sample space for three coin tosses is
$S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$.

- A = {HHH} is a simple event.
- B = {HHT, HTH, THH} is a compound event.
- C = {TTT} is a simple event.
- D = {HHH, HHT, HTH, THH} (a head shows on the first coin) is a compound event.

Mutually exclusive events are any pair with no common outcome:
- $A$ and $B$
- $A$ and $C$
- $B$ and $C$
- $C$ and $D$

Simple events: $A$ and $C$.

Compound events: $B$ and $D$.

For three coin tosses, $S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$.

(i) Two mutually exclusive events:
- A: exactly three heads = {HHH}
- B: exactly three tails = {TTT}

(ii) Three mutually exclusive and exhaustive events:
- A: no head = {TTT}
- B: exactly one head = {HTT, THT, TTH}
- C: at least two heads = {HHT, HTH, THH, HHH}

(iii) Two not mutually exclusive events:
- A: at least one head
- B: at least two heads

(iv) Two mutually exclusive but not exhaustive events:
- A: exactly one head
- B: exactly three heads

(v) Three mutually exclusive but not exhaustive events:
- A: exactly one head
- B: exactly two heads
- C: exactly three heads

These do not include the event of no head, so they are not exhaustive.

For two dice, the sample space is ordered pairs $(x,y)$.

Given:
- A: first die is even
- B: first die is odd
- C: sum $\le 5$

Then:

(i) A': first die is odd = B = $\{(1,y),(3,y),(5,y): y=1,2,3,4,5,6\}$

(ii) not B: first die is even = A

(iii) A or B: all outcomes, since the first die is either even or odd = $S$

(iv) A and B: impossible, so $\varnothing$

(v) A but not C: first die even and sum $>5$

(vi) B or C: outcomes where first die is odd or sum $\le 5$

(vii) B and C: first die odd and sum $\le 5$

(viii) A \cap B' \cap C': since $B' = A$, this is $A \cap A \cap C' = A \cap C'$, i.e. first die even and sum $>5$.

From Question 6 above, $A$ = first die even, $B$ = first die odd, $C$ = sum $\le 5$.

- (i) True: A and B cannot happen together.
- (ii) False: they are mutually exclusive, but not exhaustive as stated in the textbook wording? Actually $A\cup B=S$, so they are exhaustive too. Therefore this statement is True.
- (iii) True: $A=B'$ because $B$ is odd on the first die and $A$ is even on the first die.
- (iv) False: A and C are not mutually exclusive; e.g. $(2,2)$ belongs to both.
- (v) False: $B'=A$, so A and $B'$ are the same event, not mutually exclusive.
- (vi) False: $A'$ and $B'$ are not mutually exclusive, and together with $C$ they are not pairwise disjoint exhaustive events.

So the correct set is: (i) True, (ii) True, (iii) True, (iv) False, (v) False, (vi) False.

A valid assignment must satisfy:
1. each probability is between 0 and 1, and
2. the sum is 1.

- (a) $0.1+0.01+0.05+0.03+0.01+0.2+0.6=1$, and all are between 0 and 1, so valid.
- (b) $7\times \frac17=1$, so valid.
- (c) $0.1+0.2+0.3+0.4+0.5+0.6+0.7=2.8>1$, so invalid.
- (d) contains negative probabilities, so invalid.
- (e) sum $=\frac{1+2+3+4+5+6+15}{14}=\frac{36}{14}>1$, so invalid.

So the assignments that cannot be valid are (c), (d), and (e).

For two tosses, the sample space is $\{HH, HT, TH, TT\}$.

“At least one tail” = $\{HT, TH, TT\}$.

So
$$P(\text{at least one tail})=\frac{3}{4}.$$

The computed answer is $\frac34$; if the printed options do not include it, then the correct value is still $\frac34$.

The textbook question continues with subparts such as the probability of drawing an ace, a black card, etc. Those details are needed to answer it.

The textbook question asks for the probability of specific sums when a fair coin marked 1 and 6 and a fair die are tossed, but the exact subparts are missing.

There are $4$ men and $6$ women, so total members $=10$.

Probability of selecting a woman:
$$\frac{6}{10}=\frac{3}{5}.$$

The sample space for 4 tosses has $2^4=16$ outcomes. If $h$ heads occur, then tails $=4-h$.

Money won/lost:
$$\text{Amount}=1\cdot h-1.5(4-h)=2.5h-6.$$
For $h=0,1,2,3,4$, the amounts are:
- $h=0$: $-6$
- $h=1$: $-3.5$
- $h=2$: $-1$
- $h=3$: $1.5$
- $h=4$: $4$

So there are 5 different amounts.

With equally likely outcomes, probability of exactly $h$ heads is
$$\frac{\binom{4}{h}}{16}.$$
Thus:
- $P(-6)=\frac{\binom40}{16}=\frac{1}{16}$
- $P(-3.5)=\frac{\binom41}{16}=\frac{4}{16}=\frac14$
- $P(-1)=\frac{\binom42}{16}=\frac{6}{16}=\frac38$
- $P(1.5)=\frac{\binom43}{16}=\frac{4}{16}=\frac14$
- $P(4)=\frac{\binom44}{16}=\frac{1}{16}$

For any event $A$,
$$P(A')=1-P(A).$$
Given $P(A)=\frac{2}{11}$,
$$P(\text{not }A)=1-\frac{2}{11}=\frac{9}{11}.$$

The word ASSASSINATION has 13 letters.

Vowels: A, I, A, I, O = 5

Consonants = 13 - 5 = 8

So:
- Probability of a vowel $=\frac{5}{13}$
- Probability of a consonant $=\frac{8}{13}$

A person chooses 6 different numbers from 1 to 20. The total number of selections is
$${20\choose 6}.$$
Only one selection matches the fixed winning set.

Therefore,
$$P(\text{winning})=\frac{1}{{20\choose 6}}=\frac{1}{38760}.$$

If A and B are mutually exclusive, then
$$P(A\cup B)=P(A)+P(B).$$
So
$$P(A\text{ or }B)=\frac35+\frac15=\frac45.$$