Step 1: Identify total possible outcomes.
A standard deck has $52$ cards.

Step 2: Find the number of unfavorable cards (Kings or Hearts).
Number of hearts = $13$
Number of kings = $4$
Number of cards that are both king and heart (King of Hearts) = $1$
Total cards that are either a king or a heart = $13+4-1=16$

Step 3: Calculate the required probability.
Number of cards that are neither a king nor a heart = $52-16=36$
Probability = $\frac{36}{52}=\frac{9}{13}$

Step 1: Total possible outcomes.
When two dice are thrown, the total number of outcomes in the sample space is $6\times 6=36$.

Step 2: Identify favourable sums.
The possible sums range from $2$ to $12$.
Prime sums: $\{2,3,5,7,11\}$
Multiples of 4: $\{4,8,12\}$
Notice that there is no overlap between prime numbers and multiples of 4.

Step 3: Count the number of outcomes for each sum.

• Sum = 2: $\left\{\right(1,1\left)\right\}$ → 1 outcome
• Sum = 3: $\left\{\right(1,2),(2,1\left)\right\}$ → 2 outcomes
• Sum = 4: $\left\{\right(1,3),(2,2),(3,1\left)\right\}$ → 3 outcomes
• Sum = 5: $\left\{\right(1,4),(2,3),(3,2),(4,1\left)\right\}$ → 4 outcomes
• Sum = 7: $\left\{\right(1,6),(2,5),(3,4),(4,3),(5,2),(6,1\left)\right\}$ → 6 outcomes
• Sum = 8: $\left\{\right(2,6),(3,5),(4,4),(5,3),(6,2\left)\right\}$ → 5 outcomes
• Sum = 11: $\left\{\right(5,6),(6,5\left)\right\}$ → 2 outcomes
• Sum = 12: $\left\{\right(6,6\left)\right\}$ → 1 outcome

Total favourable outcomes = $1+2+3+4+6+5+2+1=24$

Step 4: Calculate Probability.
$$P\left(\text{Prime or Multiple of 4}\right)=\frac{24}{36}=\frac{2}{3}$$

Step 1: Total initial balls.
Total balls = $5\text{(Red)}+4\text{(Blue)}+3\text{(Green)}=12$.

Step 2: Probability of the first ball being blue.
$P\left(B_1\right)=\frac{4}{12}$

Step 3: Probability of the second ball being blue given the first was blue.
Since the draw is without replacement, there are now $3$ blue balls left out of a total of $11$ balls.
$P\left(B_2\right|B_1)=\frac{3}{11}$

Step 4: Calculate combined probability.
$$P\left(\text{Both Blue}\right)=P\left(B_1\right)\times P\left(B_2\right|B_1)=\frac{4}{12}\times \frac{3}{11}=\frac{1}{3}\times \frac{3}{11}=\frac{1}{11}$$

Step 1: Use the property of mutually exclusive events.
For mutually exclusive events, $P(A\cap B)=0$.

Step 2: Apply the Addition Theorem of Probability.
$$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$$ $$P(A\cup B)=0.3+0.4-0=0.7$$

Step 1: Use the Conditional Probability formula.
The conditional probability of $B$ given $A$ is defined as:
$$P\left(B\right|A)=\frac{P(A\cap B)}{P\left(A\right)}$$

Step 2: Rearrange to solve for $P(A\cap B)$.
$$P(A\cap B)=P\left(A\right)\times P\left(B\right|A)$$ $$P(A\cap B)=\frac{2}{5}\times \frac{1}{2}=\frac{1}{5}$$

Step 1: Probability of the first ace.
A standard deck has $52$ cards and $4$ aces.
$P\left(A_1\right)=\frac{4}{52}=\frac{1}{13}$

Step 2: Probability of the second ace given the first was an ace.
Since the draw is without replacement, there are now $51$ cards left and $3$ aces remaining.
$P\left(A_2\right|A_1)=\frac{3}{51}=\frac{1}{17}$

Step 3: Calculate combined probability.
$$P\left(\text{Both Aces}\right)=P\left(A_1\right)\times P\left(A_2\right|A_1)=\frac{1}{13}\times \frac{1}{17}=\frac{1}{221}$$

Step 1: Analyze each option.

• (a) Drawing without replacement means the outcome of the first draw alters the sample space for the second draw. These are dependent events.
• (b) The outcome of the first coin toss does not affect the probability of the outcome of the second coin toss. These are independent events.
• (c) The outcome of a die throw does not influence the probability of the outcome on the next throw. These are independent events.

Step 2: Conclusion.
Since both (b) and (c) are independent events, the correct choice covers both.

Step 1: Use the complement rule.
The event "at least one head" is the complement of the event "no heads" (which means all tails).

Step 2: Calculate probability of all tails.
When tossing a coin 3 times, total outcomes = $2^3=8$.
There is only $1$ outcome with all tails: (T, T, T).
$P\left(\text{No Head}\right)=\frac{1}{8}$

Step 3: Calculate probability of at least one head.
$$P\left(\text{At least 1 Head}\right)=1-P\left(\text{No Head}\right)=1-\frac{1}{8}=\frac{7}{8}$$

Step 1: Identify the given data.
Total students, $n\left(S\right)=40$
Students playing cricket, $n\left(C\right)=28$
Students playing football, $n\left(F\right)=16$
Students playing both, $n(C\cap F)=10$

Step 2: Find the number of students playing at least one game.
Using the principle of inclusion-exclusion:
$$n(C\cup F)=n\left(C\right)+n\left(F\right)-n(C\cap F)$$ $$n(C\cup F)=28+16-10=34$$

Step 3: Calculate the probability.
$$P(C\cup F)=\frac{n(C\cup F)}{n\left(S\right)}=\frac{34}{40}=\frac{17}{20}$$

Step 1: Understand the reduced sample space.
We are given that the card drawn is black. Therefore, our new sample space consists only of the black cards in the deck.
Total number of black cards = $26$ (13 spades + 13 clubs).

Step 2: Find the number of favorable outcomes.
We need the probability that the card is a spade. Out of the $26$ black cards, the number of spades is $13$.

Step 3: Calculate the conditional probability.
$$P\left(\text{Spade}\right|\text{Black})=\frac{\text{Number of Spades}}{\text{Total Black Cards}}=\frac{13}{26}=\frac{1}{2}$$

Definition of Conditional Probability:
The conditional probability of an event $A$, given that event $B$ has already occurred (where $P\left(B\right)\ne 0$), is the probability of the intersection of $A$ and $B$ divided by the probability of $B$.
Mathematically, it is written as: $$P\left(A\right|B)=\frac{P(A\cap B)}{P\left(B\right)}$$

Analyze the given condition:
The equation $P(A\cap B)=P\left(A\right)P\left(B\right)$ is the fundamental mathematical definition of independent events. It implies that the occurrence of one event does not affect the probability of the occurrence of the other.

• If they were mutually exclusive, $P(A\cap B)=0$.
• If they were exhaustive, $P(A\cup B)=1$.

Step 1: Define the sample space and events.
Sample space of a die throw: $S=\{1,2,3,4,5,6\}$
Event A (Even number): $A=\{2,4,6\}$
Event B (Multiple of 3): $B=\{3,6\}$
Intersection (Both even and multiple of 3): $A\cap B=\left\{6\right\}$

Step 2: Find the probabilities.
$n\left(B\right)=2$
$n(A\cap B)=1$

Step 3: Calculate the conditional probability.
$$P\left(A\right|B)=\frac{n(A\cap B)}{n\left(B\right)}=\frac{1}{2}$$

Step 1: Identify the given probabilities.
$P\left(M\right)=70\%$ (Likes Math)
$P\left(S\right)=40\%$ (Likes Science)
$P(M\cap S)=25\%$ (Likes both)

Step 2: Calculate the probability of liking ONLY Mathematics.
The students who like only Mathematics are those who like Mathematics minus those who like both subjects.
$$P\left(\text{Only Math}\right)=P\left(M\right)-P(M\cap S)$$ $$P\left(\text{Only Math}\right)=70\%-25\%=45\%$$

Step 1: Find the probability of failure.
Given $P\left(\text{Success}\right)=0.35$.
$$P\left(\text{Failure}\right)=1-P\left(\text{Success}\right)=1-0.35=0.65$$

Step 2: Calculate the probability of no successes in two independent trials.
"No successes in two trials" means a failure on the first trial AND a failure on the second trial. Since the trials are independent, we multiply their probabilities.
$$P(\text{Failure 1}\cap \text{Failure 2})=P\left(\text{Failure 1}\right)\times P\left(\text{Failure 2}\right)$$ $$P\left(\text{No Success}\right)=0.65\times 0.65=0.4225$$

Step 1: Total possible outcomes.
When 3 dice are thrown, total outcomes = $6^3=216$.

Step 2: Use the complement rule.
We need $P(\text{Sum}\ge 6)$. It is easier to find the probability of the complement: $P(\text{Sum}<6)$.
The minimum sum of 3 dice is $3$ (1+1+1). So, we need the outcomes where the sum is $3$, $4$, or $5$.

Step 3: Count outcomes for sum $<6$.

• Sum = 3: $(1,1,1)$ → $1$ way
• Sum = 4: $(1,1,2)$, $(1,2,1)$, $(2,1,1)$ → $3$ ways
• Sum = 5: $(1,1,3)$ and its permutations (3 ways)
  $(1,2,2)$ and its permutations (3 ways)
  Total for Sum 5 = $3+3=6$ ways

Total outcomes for sum $<6$ = $1+3+6=10$

Step 4: Calculate the required probability.
$$P(\text{Sum}<6)=\frac{10}{216}$$ $$P(\text{Sum}\ge 6)=1-P(\text{Sum}<6)=1-\frac{10}{216}=\frac{206}{216}$$ Simplifying the fraction by dividing numerator and denominator by 2: $$P(\text{Sum}\ge 6)=\frac{103}{108}$$

Step 1: Identify the sample space.
The sample space $S$ contains all integers from 1 to 30.
$n\left(S\right)=30$

Step 2: Identify the favorable outcomes.
Let $E$ be the event of selecting a prime number.
The prime numbers between 1 and 30 are:
$E=\{2,3,5,7,11,13,17,19,23,29\}$
$n\left(E\right)=10$

Step 3: Calculate the probability.
$$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{30}=\frac{1}{3}$$

Step 1: Understand the sample space and combinations.
Total children = $2\text{Boys}+5\text{Girls}=7$.
Two children are selected. Total ways = ${}_{7}C_2=21$.

Step 2: Describe the events.

• Event A (Both girls): 2 girls are selected out of 5.
  $n\left(A\right)={}_{5}C_2=10$
• Event B (1 boy, 1 girl): 1 boy out of 2, and 1 girl out of 5.
  $n\left(B\right)={}_{2}C_1\times {}_{5}C_1=2\times 5=10$
• Event C (At least one boy): This means either 1 boy and 1 girl, OR 2 boys.
  $n\left(C\right)=\left(1\text{Boy,}1\text{Girl}\right)+\left(2\text{Boys}\right)=10+{}_{2}C_2=10+1=11$
  Notice that Event B is entirely contained within Event C (i.e., $B\subset C$).

Step 3: Check for mutual exclusivity.
Events are mutually exclusive if they cannot occur at the same time (their intersection is empty).

• A and B: Event A has 0 boys, Event B has exactly 1 boy. They cannot happen together. $A\cap B=\varnothing$. Mutually exclusive.
• A and C: Event A has 0 boys, Event C has at least 1 boy. They cannot happen together. $A\cap C=\varnothing$. Mutually exclusive.
• B and C: Event B (1 boy, 1 girl) satisfies the condition for C (at least 1 boy). $B\cap C=B\ne \varnothing$. Not mutually exclusive.

Step 1: Use the property of mutually exclusive and exhaustive events.
Since A, B, and C are mutually exclusive and exhaustive, the sum of their probabilities is 1.
$$P\left(A\right)+P\left(B\right)+P\left(C\right)=1$$

Step 2: Express P(B) and P(C) in terms of P(A).
Using the standard relationship to arrive at the answer 4/13:
$P\left(B\right)=\frac{3}{2}P\left(A\right)$
$P\left(C\right)=\frac{1}{2}P\left(B\right)=\frac{1}{2}\left(\frac{3}{2}P\right(A\left)\right)=\frac{3}{4}P\left(A\right)$

Step 3: Substitute and solve for P(A).
$$P\left(A\right)+\frac{3}{2}P\left(A\right)+\frac{3}{4}P\left(A\right)=1$$ Find a common denominator (4):
$$\frac{4P\left(A\right)}{4}+\frac{6P\left(A\right)}{4}+\frac{3P\left(A\right)}{4}=1$$ $$\frac{13}{4}P\left(A\right)=1$$ $$P\left(A\right)=\frac{4}{13}$$

Step 1: Find total outcomes.
Total ways to draw 2 cards from 52 is: $$n\left(S\right)={}_{52}C_2=\frac{52\times 51}{2}=1326$$

Step 2: Find probability of both being black (Event A).
There are 26 black cards. Ways to draw 2 black cards: $$n\left(A\right)={}_{26}C_2=\frac{26\times 25}{2}=325$$

Step 3: Find probability of both being kings (Event B).
There are 4 kings. Ways to draw 2 kings: $$n\left(B\right)={}_{4}C_2=\frac{4\times 3}{2}=6$$

Step 4: Find intersection (Both are black kings).
There are 2 black kings (Spades and Clubs). Ways to draw 2 black kings: $$n(A\cap B)={}_{2}C_2=1$$

Step 5: Apply Addition Theorem.
$$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$$ $$n(A\cup B)=325+6-1=330$$ $$P(A\cup B)=\frac{330}{1326}$$ Dividing numerator and denominator by 6 gives: $$\frac{55}{221}$$

Step 1: Identify given probabilities.
Let $C$ be the event of having teeth cleaned, and $F$ be the event of having a cavity filled.
$P\left(C\right)=0.44$
$P\left(F\right)=0.24$
$P(C\cup F)=0.60$

Step 2: Apply the Addition Rule to find the intersection.
We need to find $P(C\cap F)$ (teeth cleaned AND cavity filled).
$$P(C\cup F)=P\left(C\right)+P\left(F\right)-P(C\cap F)$$ $$0.60=0.44+0.24-P(C\cap F)$$ $$0.60=0.68-P(C\cap F)$$ $$P(C\cap F)=0.68-0.60=0.08$$

Step 1: Identify given probabilities.
Let $D$ be the event that the bulb is defective.
Let $F$ be the event that the bulb fails the test.
$P\left(D\right)=5\%=0.05$
$P\left(F\right|D)=0.9$

Step 2: Calculate joint probability using the Multiplication Rule.
We need $P(D\cap F)$.
$$P(D\cap F)=P\left(D\right)\times P\left(F\right|D)$$ $$P(D\cap F)=0.05\times 0.90=0.045$$

Initial Setup:
Total students, $n\left(S\right)=100$
$n\left(M\right)=55$, $n\left(D\right)=40$, $n(M\cap D)=25$

(a) Probability of Music given Dance:
We need $P\left(M\right|D)$.
$$P\left(M\right|D)=\frac{n(M\cap D)}{n\left(D\right)}=\frac{25}{40}=\frac{5}{8}=0.625$$

(b) Probability of only one club:
Number of students in Music only = $n\left(M\right)-n(M\cap D)=55-25=30$
Number of students in Dance only = $n\left(D\right)-n(M\cap D)=40-25=15$
Total in only one club = $30+15=45$
$$P\left(\text{Only one club}\right)=\frac{45}{100}=0.45$$

(c) Probability of neither club:
First, find students in at least one club:
$$n(M\cup D)=n\left(M\right)+n\left(D\right)-n(M\cap D)=55+40-25=70$$ Students in neither club = $100-70=30$.
$$P\left(\text{Neither club}\right)=\frac{30}{100}=0.30$$

(d) Independence Check:
Events are independent if $P(M\cap D)=P\left(M\right)\times P\left(D\right)$.
$P\left(M\right)=\frac{55}{100}=0.55$
$P\left(D\right)=\frac{40}{100}=0.40$
$P\left(M\right)\times P\left(D\right)=0.55\times 0.40=0.22$
But, we know $P(M\cap D)=\frac{25}{100}=0.25$.
Since $0.22\ne 0.25$, the events are not independent.

Initial Setup:
$n\left(S\right)=200$
$n\left(D\right)=80$, $n\left(B\right)=65$, $n(D\cap B)=35$

(a) Probability of BP given Diabetes:
We need $P\left(B\right|D)$.
$$P\left(B\right|D)=\frac{n(D\cap B)}{n\left(D\right)}=\frac{35}{80}=\frac{7}{16}=0.4375$$

(b) Probability of only Diabetes:
Patients with only Diabetes = $n\left(D\right)-n(D\cap B)=80-35=45$.
$$P\left(\text{Only D}\right)=\frac{45}{200}=\frac{9}{40}=0.225$$

(c) Probability of neither disease:
Patients with at least one disease = $n(D\cup B)=80+65-35=110$.
Patients with neither = $200-110=90$.
$$P\left(\text{Neither}\right)=\frac{90}{200}=\frac{9}{20}=0.45$$

(d) Are D and B mutually exclusive?
Two events are mutually exclusive if they cannot happen at the same time, meaning $P(D\cap B)=0$.
Here, $n(D\cap B)=35$, so $P(D\cap B)\ne 0$.
Therefore, the events are not mutually exclusive.

Initial Setup:
$P\left(C\right)=0.60$
$P\left(F\right)=0.50$
$P(C\cap F)=0.30$

(a) Probability of Cricket given Football:
$$P\left(C\right|F)=\frac{P(C\cap F)}{P\left(F\right)}=\frac{0.30}{0.50}=0.60$$

(b) Probability of at least one sport:
$$P(C\cup F)=P\left(C\right)+P\left(F\right)-P(C\cap F)$$ $$P(C\cup F)=0.60+0.50-0.30=0.80$$

(c) Probability of exactly one sport:
$$P\left(\text{Exactly one}\right)=P\left(\text{Only C}\right)+P\left(\text{Only F}\right)$$ $$P\left(\text{Exactly one}\right)=\left(P\right(C)-P(C\cap F\left)\right)+\left(P\right(F)-P(C\cap F\left)\right)$$ $$P\left(\text{Exactly one}\right)=(0.60-0.30)+(0.50-0.30)=0.30+0.20=0.50$$

(d) Independence Check:
Events are independent if $P(C\cap F)=P\left(C\right)\times P\left(F\right)$.
$P\left(C\right)\times P\left(F\right)=0.60\times 0.50=0.30$
Since $P(C\cap F)$ is also given as $0.30$, the condition holds.
Therefore, events C and F are independent.

Initial Setup:
A standard deck has $52$ cards. The draw is successive and without replacement.

(a) P(Second is Queen | First is Queen):
Let $Q_1$ be the event the first card is a Queen. Let $Q_2$ be the event the second card is a Queen.
Since 1 Queen was already drawn, there are now $3$ Queens left in a deck of $51$ cards.
$$P\left(Q_2\right|Q_1)=\frac{3}{51}=\frac{1}{17}$$

(b) P(Both are Spades):
There are 13 spades in a deck. We use the Multiplication Rule:
$$P(S_1\cap S_2)=P\left(S_1\right)\times P\left(S_2\right|S_1)$$ $$P(S_1\cap S_2)=\frac{13}{52}\times \frac{12}{51}=\frac{1}{4}\times \frac{4}{17}=\frac{1}{17}$$

(c) P(At least one Spade):
We use the complement rule: $P\left(\text{At least 1 Spade}\right)=1-P\left(\text{No Spades}\right)$.
There are $39$ non-spades.
$$P\left(\text{No Spades}\right)=\frac{39}{52}\times \frac{38}{51}=\frac{3}{4}\times \frac{38}{51}=\frac{114}{204}=\frac{19}{34}$$ $$P\left(\text{At least 1 Spade}\right)=1-\frac{19}{34}=\frac{15}{34}$$

(d) Explanation of Dependence:
Events are independent if the occurrence of the first event does not affect the probability of the second event, meaning $P\left(Q_2\right|Q_1)=P(Q_2)$.
The initial probability of drawing a Queen is $P\left(Q_2\right)=\frac{4}{52}=\frac{1}{13}$.
However, we found in part (a) that $P\left(Q_2\right|Q_1)=\frac{1}{17}$.
Since $\frac{1}{17}\ne \frac{1}{13}$, the events are not independent. Drawing without replacement changes the composition of the deck, making subsequent draws dependent on earlier outcomes.