Given: A bag contains only lemon-flavoured candies. Malini takes out one candy at random.

To Find: The probability that the candy taken out is an orange-flavoured candy.

Step 1: Defining the Sample Space and Events

Let $S$ be the sample space, which consists of all possible outcomes of taking out a candy from the bag. Since the bag contains only lemon-flavoured candies, every candy in the bag is a lemon-flavoured candy.

Let $n(S)$ be the total number of candies in the bag. Let $L$ be the number of lemon-flavoured candies. Thus, $n(S) = L$.

Step 2: Defining the Event of Interest

Let $E$ be the event of taking out an orange-flavoured candy. Since the bag contains only lemon-flavoured candies, there are no orange-flavoured candies present in the bag.

Therefore, the number of favourable outcomes for event $E$, denoted as $n(E)$, is $0$.

Step 3: Applying the Probability Formula

The theoretical probability of an event $E$ is defined by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$

Step 4: Calculation

Substituting the values identified in the previous steps:

$P(E) = \frac{0}{L}$

[Since $L > 0$, as there is at least one candy in the bag to be taken out]

$P(E) = 0$

Justification:

An event that has no chance of occurring is called an impossible event. The probability of an impossible event is always $0$. Since there are no orange-flavoured candies in the bag, it is impossible to draw one.

Final Answer: The probability that she takes out an orange flavoured candy is 0.

Given: A game of chance involves a spinner with 8 equal sectors labeled with the numbers $\{1, 2, 3, 4, 5, 6, 7, 8\}$. The arrow is equally likely to stop at any of these numbers.

To Find: The probability that the arrow points at the number $8$.

Visual Representation:

Step 1: Define the Sample Space
The sample space $S$ is the set of all possible outcomes of the experiment. Since the spinner has 8 numbers, the sample space is:
$S = \{1, 2, 3, 4, 5, 6, 7, 8\}$
The total number of possible outcomes, denoted by $n(S)$, is $8$.

Step 2: Define the Favorable Event
Let $E$ be the event that the arrow points at the number $8$.
The set of favorable outcomes is $E = \{8\}$.
The number of favorable outcomes, denoted by $n(E)$, is $1$.

Step 3: Apply the Probability Formula
The theoretical probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$
$P(E) = \frac{n(E)}{n(S)}$

Step 4: Calculation
Substituting the values obtained in Step 1 and Step 2 into the formula:
$P(E) = \frac{1}{8}$

Final Answer: The probability that the arrow will point at 8 is $\frac{1}{8}$.

Given: An event $E$ associated with a random experiment and its corresponding complementary event 'not $E$', denoted as $E^c$ or $\overline{E}$.

To Find: The sum of the probability of event $E$ and the probability of the event 'not $E$'.

Step 1: Defining the Probability of an Event
Let $S$ be the sample space of a random experiment. Let $E$ be an event such that $E \subseteq S$. The probability of event $E$, denoted by $P(E)$, is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes in the sample space $S$.
$P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes in } S}$

Step 2: Defining the Complementary Event
The event 'not $E$', denoted as $\overline{E}$, consists of all outcomes in the sample space $S$ that are not in $E$. Mathematically, $\overline{E} = S \setminus E$.
Since $E$ and $\overline{E}$ are mutually exclusive (they cannot occur simultaneously) and exhaustive (their union covers the entire sample space $S$), we have:
$E \cup \overline{E} = S$
$E \cap \overline{E} = \emptyset$

Step 3: Applying the Axiom of Probability
According to the axiomatic definition of probability, the sum of the probabilities of all elementary events in a sample space is equal to $1$.
Since $E$ and $\overline{E}$ partition the sample space $S$, the sum of their probabilities must equal the probability of the sure event (the sample space itself).
$P(E) + P(\overline{E}) = P(S)$

Step 4: Final Calculation
The probability of the sure event (the entire sample space $S$) is always $1$.
$P(S) = 1$
Therefore:
$P(E) + P(\text{not } E) = 1$

Final Answer: 1

Given: A game involves tossing a fair one-rupee coin 3 times. Hanif wins if all three tosses result in the same outcome (HHH or TTT). Hanif loses if the outcomes are not all the same.

To Find: The probability that Hanif will lose the game.

Step 1: Determining the Sample Space

When a coin is tossed once, there are 2 possible outcomes: Head (H) or Tail (T). When a coin is tossed 3 times, the total number of possible outcomes is $2^3 = 2 \times 2 \times 2 = 8$.

Let $S$ be the sample space representing all possible outcomes:

$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$

The total number of elementary events, $n(S) = 8$.

Step 2: Identifying Winning Outcomes

Hanif wins if all tosses result in the same outcome. These outcomes are:

Winning outcomes = $\{HHH, TTT\}$

Number of winning outcomes, $n(W) = 2$.

Step 3: Identifying Losing Outcomes

Hanif loses if the outcome is not one of the winning outcomes. These outcomes are:

Losing outcomes = $\{HHT, HTH, HTT, THH, THT, TTH\}$

Number of losing outcomes, $n(L) = n(S) - n(W) = 8 - 2 = 6$.

Step 4: Calculating the Probability of Losing

The probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Let $L$ be the event that Hanif loses the game.

$P(L) = \frac{n(L)}{n(S)}$

$P(L) = \frac{6}{8}$

Step 5: Simplifying the Fraction

To simplify $\frac{6}{8}$, we divide both the numerator and the denominator by their greatest common divisor, which is 2:

$P(L) = \frac{6 \div 2}{8 \div 2} = \frac{3}{4}$

Alternatively, using the complement rule:

$P(W) = \frac{n(W)}{n(S)} = \frac{2}{8} = \frac{1}{4}$

$P(L) = 1 - P(W)$ [Since the sum of probabilities of complementary events is 1]

$P(L) = 1 - \frac{1}{4} = \frac{3}{4}$

Final Answer: The probability that Hanif will lose the game is $\frac{3}{4}$ or $0.75$.

Given: An experiment where a player attempts to shoot a basketball, resulting in either a "shot" (success) or a "miss" (failure).

To Find: Determine whether the outcomes of this experiment are "equally likely" and provide a logical explanation.

Definition of Equally Likely Outcomes:
Outcomes of an experiment are said to be equally likely if each outcome has the same probability of occurring. If an experiment has $n$ outcomes, each outcome is equally likely if the probability of each is $\frac{1}{n}$.

Step 1: Identifying the Sample Space
Let $S$ be the sample space of the experiment.
The possible outcomes are:
$E_1$: The player shoots the ball (Success).
$E_2$: The player misses the shot (Failure).
Thus, $S = \{E_1, E_2\}$.

Step 2: Analyzing the Nature of the Experiment
The probability of a basketball player making or missing a shot depends on several factors, including:
1. The skill level of the player.
2. The distance from the basket.
3. The defensive pressure.
4. The physical and mental state of the player.

Step 3: Evaluating the Likelihood
Let $P(E_1)$ be the probability of shooting the ball successfully and $P(E_2)$ be the probability of missing the shot.
For the outcomes to be equally likely, we must have $P(E_1) = P(E_2) = 0.5$.
However, in a real-world scenario, a professional player has a high probability of success ($P(E_1) > 0.5$), while a novice player may have a low probability of success ($P(E_1) < 0.5$). Since the probability is dependent on the player's ability and external conditions, it is not fixed at $0.5$ for every player or every attempt.

Step 4: Conclusion
Since the probability of hitting the basket is not necessarily equal to the probability of missing the basket, the outcomes are not equally likely.

Final Answer: The outcomes are not equally likely because the probability of shooting or missing depends on the player's skill and various other factors, and is not inherently fixed at $50\%$ for each outcome.

Given: A standard six-faced die is thrown. The possible outcomes are categorized into two groups: odd numbers and even numbers.

To Determine: Whether the argument "the probability of getting an odd number is $\frac{1}{2}$" is correct, and to provide a rigorous mathematical justification.

Visual Representation of the Sample Space:

Step 1: Defining the Sample Space

When a fair six-faced die is thrown, the set of all possible outcomes (Sample Space $S$) is given by:
$S = \{1, 2, 3, 4, 5, 6\}$
The total number of elementary outcomes, denoted by $n(S)$, is $6$.

Step 2: Identifying the Event of Interest

Let $E$ be the event of getting an odd number. The odd numbers present on a standard die are $1, 3,$ and $5$.
Therefore, $E = \{1, 3, 5\}$.
The number of favorable outcomes for event $E$, denoted by $n(E)$, is $3$.

Step 3: Applying the Probability Formula

The theoretical probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes:
$P(E) = \frac{n(E)}{n(S)}$
Substituting the values identified in the previous steps:
$P(E) = \frac{3}{6}$

Step 4: Simplifying the Fraction

By dividing both the numerator and the denominator by their greatest common divisor, which is $3$:
$P(E) = \frac{3 \div 3}{6 \div 3} = \frac{1}{2}$

Step 5: Justification of the Argument

The argument states that because there are two categories (odd and even), the probability is $\frac{1}{2}$. While the result $\frac{1}{2}$ is numerically correct, the reasoning is valid only because each of the six outcomes $\{1, 2, 3, 4, 5, 6\}$ is equally likely. Since there are exactly three odd numbers and three even numbers, the probability of selecting an odd number is indeed $\frac{3}{6} = \frac{1}{2}$.

Final Answer: The argument is correct. The probability of getting an odd number is $\frac{1}{2}$ because there are 3 odd outcomes out of 6 equally likely total outcomes.

Given: A game of chance involves a spinner with numbers $1, 2, 3, 4, 5, 6, 7, 8$. The outcomes are equally likely.

To Find: The probability that the arrow points at an odd number.

Step 1: Define the Sample Space
The sample space $S$ consists of all possible outcomes of the spinner.
$S = \{1, 2, 3, 4, 5, 6, 7, 8\}$
The total number of possible outcomes, denoted by $n(S)$, is $8$.

Step 2: Identify the Favorable Outcomes
Let $E$ be the event of getting an odd number.
An odd number is an integer that is not divisible by $2$.
From the set $S$, the odd numbers are $\{1, 3, 5, 7\}$.
The number of favorable outcomes, denoted by $n(E)$, is $4$.

Step 3: Apply the Probability Formula
The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.
Formula: $P(E) = \frac{n(E)}{n(S)}$
[Since all outcomes are equally likely]

Step 4: Calculation
Substitute the values into the formula:
$P(E) = \frac{4}{8}$
Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is $4$:
$P(E) = \frac{4 \div 4}{8 \div 4} = \frac{1}{2}$

Final Answer: The probability that the arrow will point at an odd number is $\frac{1}{2}$ or $0.5$.

Given: A fair, standard six-faced die is thrown once. The possible outcomes on the faces of the die are $\{1, 2, 3, 4, 5, 6\}$.

To find: The probability of getting a number lying between 2 and 6.

Visual Representation of the Sample Space:

Step 1: Define the Sample Space ($S$)
The sample space $S$ consists of all possible outcomes when a die is thrown:
$S = \{1, 2, 3, 4, 5, 6\}$
The total number of possible outcomes, denoted by $n(S)$, is $6$.

Step 2: Define the Event ($E$)
Let $E$ be the event of getting a number lying between 2 and 6. The numbers strictly between 2 and 6 are 3, 4, and 5.
$E = \{3, 4, 5\}$
The number of favorable outcomes, denoted by $n(E)$, is $3$.

Step 3: Apply the Probability Formula
The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes:
$P(E) = \frac{n(E)}{n(S)}$
[Using the classical definition of probability]

Step 4: Calculation
Substitute the values identified in Step 1 and Step 2 into the formula:
$P(E) = \frac{3}{6}$
Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
$P(E) = \frac{3 \div 3}{6 \div 3} = \frac{1}{2}$

Final Answer: The probability of getting a number lying between 2 and 6 is $\frac{1}{2}$.

Given: A bag contains only lemon-flavoured candies. Malini takes out one candy from the bag without looking.

To Find: The probability that the candy taken out is a lemon-flavoured candy.

Step 1: Defining the Sample Space and Event

Let the total number of candies in the bag be $n$. Since the bag contains only lemon-flavoured candies, the number of lemon-flavoured candies in the bag is also $n$.

Let $E$ be the event of taking out a lemon-flavoured candy.

Step 2: Applying the Probability Formula

The theoretical probability of an event $E$, denoted by $P(E)$, is defined by the ratio of the number of favourable outcomes to the total number of possible outcomes:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

Step 3: Calculating the Probability

In this scenario:

• The number of favourable outcomes (taking out a lemon-flavoured candy) = $n$
• The total number of possible outcomes (total candies in the bag) = $n$

Substituting these values into the formula:

$P(E) = \frac{n}{n}$

$P(E) = 1$

Step 4: Theoretical Justification

[Since the event of picking a lemon-flavoured candy is a sure event or certain event because the bag contains no other types of candies, the probability must be 1.]

Final Answer: The probability that she takes out a lemon-flavoured candy is 1.

Given: Two coins are tossed simultaneously. The argument states that there are three possible outcomes: two heads (HH), two tails (TT), or one of each (HT or TH). Based on this, it claims the probability of each outcome is $\frac{1}{3}$.

To Find: Determine if the argument is correct or incorrect and provide a mathematical justification.

Step 1: Identifying the Sample Space
When two fair coins are tossed simultaneously, each coin has two possible outcomes: Head ($H$) or Tail ($T$). Let the outcomes of the first coin be $C_1$ and the second coin be $C_2$. The total number of possible outcomes is determined by the Fundamental Counting Principle: $2 \times 2 = 4$.
The sample space $S$ is the set of all possible elementary events:
$S = \{HH, HT, TH, TT\}$
Where:
$HH$ = Head on both coins
$HT$ = Head on the first coin, Tail on the second
$TH$ = Tail on the first coin, Head on the second
$TT$ = Tail on both coins

Step 2: Analyzing the Argument's Outcomes
The argument suggests three outcomes: "two heads", "two tails", or "one of each".
Let us map these to our sample space $S$:
- Two heads: $\{HH\}$ (1 outcome)
- Two tails: $\{TT\}$ (1 outcome)
- One of each: $\{HT, TH\}$ (2 outcomes)

Step 3: Calculating Probabilities
Using the classical definition of probability, $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
The total number of outcomes in the sample space is $n(S) = 4$.

Calculating the probability for each category defined in the argument:

1. Probability of two heads ($P_{HH}$):
$P(HH) = \frac{1}{4}$

2. Probability of two tails ($P_{TT}$):
$P(TT) = \frac{1}{4}$

3. Probability of one of each ($P_{one\_of\_each}$):
$P(HT \text{ or } TH) = \frac{2}{4} = \frac{1}{2}$

Step 4: Comparison and Conclusion
The argument claims that the probability for each of the three outcomes is $\frac{1}{3}$. However, our calculations show:
$P(HH) = \frac{1}{4} \neq \frac{1}{3}$
$P(TT) = \frac{1}{4} \neq \frac{1}{3}$
$P(one\_of\_each) = \frac{1}{2} \neq \frac{1}{3}$

The error in the argument lies in the assumption that the three outcomes are "equally likely". In probability theory, the classical formula $P(E) = \frac{1}{n}$ only applies when all elementary events in the sample space are equally likely. Since the event "one of each" consists of two distinct elementary outcomes ($HT$ and $TH$), it is twice as likely to occur as "two heads" or "two tails".

Final Answer: The argument is incorrect. The outcomes are not equally likely; the probability of getting two heads is $\frac{1}{4}$, two tails is $\frac{1}{4}$, and one of each is $\frac{1}{2}$.

Given: A fair six-faced die is thrown twice. The possible outcomes for each throw are $\{1, 2, 3, 4, 5, 6\}$.

To Find: The probability that the number 5 will not come up in either of the two throws.

Step 1: Determining the Total Number of Possible Outcomes

When a die is thrown once, there are $6$ possible outcomes. When a die is thrown twice, the total number of outcomes is calculated by the product of the outcomes of each throw [Fundamental Counting Principle].

Total outcomes = $6 \times 6 = 36$.

The sample space $S$ consists of all ordered pairs $(a, b)$ where $a$ is the result of the first throw and $b$ is the result of the second throw:

$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),$

$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),$

$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),$

$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),$

$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),$

$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

Step 2: Identifying Favorable Outcomes

Let $E$ be the event that 5 does not come up in either throw. This means that for any outcome $(a, b)$, $a \neq 5$ and $b \neq 5$.

If $a \neq 5$, then $a \in \{1, 2, 3, 4, 6\}$ (5 possibilities).

If $b \neq 5$, then $b \in \{1, 2, 3, 4, 6\}$ (5 possibilities).

The number of favorable outcomes $n(E)$ is the product of the number of choices for the first throw and the second throw:

$n(E) = 5 \times 5 = 25$.

Step 3: Calculating the Probability

The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes [Classical Definition of Probability].

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(E) = \frac{n(E)}{n(S)}$

$P(E) = \frac{25}{36}$

Alternative Method (Using Complementary Events):

Let $A$ be the event that 5 comes up at least once. The outcomes where 5 appears are:

$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)$.

Counting these, we find $n(A) = 11$.

$P(A) = \frac{11}{36}$.

Since the event "5 does not come up" is the complement of event $A$ (denoted as $A'$), we use the property $P(A') = 1 - P(A)$:

$P(A') = 1 - \frac{11}{36} = \frac{36 - 11}{36} = \frac{25}{36}$.

Final Answer: The probability that 5 will not come up either time is $\frac{25}{36}$.

Given: A fair six-faced die is thrown once. The possible outcomes are the integers from 1 to 6.

To Find: The probability of getting a prime number.

Visual Representation of the Sample Space:

Step 1: Define the Sample Space ($S$)
When a standard die is thrown, the set of all possible outcomes is:
$S = \{1, 2, 3, 4, 5, 6\}$
The total number of possible outcomes, denoted as $n(S)$, is:
$n(S) = 6$ [Since there are 6 faces on a standard die]

Step 2: Identify the Favorable Outcomes ($E$)
Let $E$ be the event of getting a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
Analyzing the set $S = \{1, 2, 3, 4, 5, 6\}$:
- 1 is neither prime nor composite.
- 2 is a prime number.
- 3 is a prime number.
- 4 is a composite number ($2 \times 2$).
- 5 is a prime number.
- 6 is a composite number ($2 \times 3$).
Thus, the set of favorable outcomes is $E = \{2, 3, 5\}$.
The number of favorable outcomes, denoted as $n(E)$, is:
$n(E) = 3$

Step 3: Apply the Probability Formula
The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes:
$P(E) = \frac{n(E)}{n(S)}$ [Definition of Theoretical Probability]

Step 4: Calculation
Substitute the values identified in Step 1 and Step 2 into the formula:
$P(E) = \frac{3}{6}$
Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
$P(E) = \frac{3 \div 3}{6 \div 3} = \frac{1}{2}$

Final Answer: The probability of getting a prime number is $\frac{1}{2}$.

Given: The definition of the probability of an event $E$, denoted as $P(E)$, in the context of classical probability theory.

To Find: The lower and upper bounds for the probability of any event $E$.

Step 1: Understanding the definition of an event
In probability theory, an event $E$ is a subset of the sample space $S$. The probability of an event is a measure of the likelihood that the event will occur. By definition, the probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space.

Step 2: Analyzing the Impossible Event
An impossible event is an event that cannot occur. The number of favorable outcomes for an impossible event is $0$.
Therefore, $P(\text{Impossible Event}) = \frac{0}{n(S)} = 0$.
This establishes the minimum possible value for any probability, which is $0$.

Step 3: Analyzing the Sure Event
A sure event (or certain event) is an event that is guaranteed to occur. The number of favorable outcomes for a sure event is equal to the total number of outcomes in the sample space, $n(S)$.
Therefore, $P(\text{Sure Event}) = \frac{n(S)}{n(S)} = 1$.
This establishes the maximum possible value for any probability, which is $1$.

Step 4: Formulating the Inequality
Since the number of favorable outcomes for any event $E$ must be greater than or equal to $0$ and cannot exceed the total number of outcomes in the sample space $n(S)$, we have the following inequality:
$0 \leq \text{Number of favorable outcomes} \leq n(S)$
Dividing the entire inequality by $n(S)$ (where $n(S) > 0$):
$\frac{0}{n(S)} \leq \frac{\text{Number of favorable outcomes}}{n(S)} \leq \frac{n(S)}{n(S)}$
$0 \leq P(E) \leq 1$

Step 5: Conclusion
Based on the derivation above, the probability of an event is always greater than or equal to $0$ and less than or equal to $1$.

Final Answer: The probability of an event is greater than or equal to 0 and less than or equal to 1.

Given:

A group of 3 students is considered. The probability that 2 students do not have the same birthday is given as $P(\text{not same birthday}) = 0.992$.

To find:

The probability that the 2 students have the same birthday, denoted as $P(\text{same birthday})$.

Step 1: Identifying the relationship between complementary events

In probability theory, for any event $E$, the event "not $E$" (denoted as $\overline{E}$) represents the complement of event $E$. The sum of the probability of an event occurring and the probability of the event not occurring is always equal to 1.

Formula: $P(E) + P(\overline{E}) = 1$

[Since the sum of probabilities of all elementary events in a sample space is 1]

Step 2: Defining the variables

Let $E$ be the event that 2 students have the same birthday.

Let $\overline{E}$ be the event that 2 students do not have the same birthday.

Given: $P(\overline{E}) = 0.992$

Step 3: Substituting the values into the formula

Using the identity $P(E) + P(\overline{E}) = 1$:

$P(E) + 0.992 = 1$

Step 4: Solving for $P(E)$

To isolate $P(E)$, subtract $0.992$ from both sides of the equation:

$P(E) = 1 - 0.992$

[Performing the subtraction: $1.000 - 0.992 = 0.008$]

$P(E) = 0.008$

Final Answer: The probability that the 2 students have the same birthday is 0.008.

Given: An experiment where a driver attempts to start a car. The possible outcomes are defined as:
1. The car starts.
2. The car does not start.

To Find: Determine whether the outcomes of this experiment are "equally likely" and provide a logical explanation.

Definition: In probability theory, outcomes of an experiment are said to be equally likely if each outcome has the same probability of occurring. That is, if an experiment has $n$ possible outcomes, each outcome must have a probability of $\frac{1}{n}$.

Step 1: Analyzing the nature of the experiment
The experiment involves a mechanical process (starting a car). The outcome depends on various factors such as:
a) The condition of the car's battery.
b) The condition of the fuel system.
c) The condition of the ignition system.
d) The ambient temperature and maintenance history.

Step 2: Evaluating the probability of outcomes
Let $E_1$ be the event that the car starts.
Let $E_2$ be the event that the car does not start.
For the outcomes to be equally likely, we would require $P(E_1) = P(E_2) = 0.5$.

Step 3: Logical Deduction
In a real-world scenario, a car is designed to start. If the car is in good working condition, the probability of it starting ($P(E_1)$) is significantly higher than the probability of it not starting ($P(E_2)$). Conversely, if the car is broken, the probability of it not starting ($P(E_2)$) is significantly higher. Since the probability of the car starting is not necessarily equal to the probability of it not starting, the outcomes are dependent on external conditions rather than being inherently balanced.

Step 4: Conclusion
Since the likelihood of the car starting is not fixed at 50% and varies based on the mechanical state of the vehicle, the outcomes are not equally likely.

Final Answer: The outcomes are not equally likely because the probability of the car starting depends on various mechanical factors and is not necessarily equal to the probability of the car not starting.

Given: A well-shuffled deck of $52$ playing cards.

To find: The probability of drawing a card that is a spade.

Visual Representation of Card Suits:

Step 1: Define the Sample Space

The total number of possible outcomes when drawing one card from a well-shuffled deck is the total number of cards in the deck.

Let $n(S)$ be the total number of outcomes.

$n(S) = 52$ [Since a standard deck contains 52 cards]

Step 2: Identify the Favorable Outcomes

A standard deck of cards is divided into four suits: Spades, Hearts, Diamonds, and Clubs. Each suit contains exactly $13$ cards.

Let $E$ be the event of drawing a spade.

The number of favorable outcomes $n(E)$ is the number of spade cards in the deck.

$n(E) = 13$ [Since there are 13 cards of the spade suit]

Step 3: Apply the Probability Formula

The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.

$P(E) = \frac{n(E)}{n(S)}$ [Definition of theoretical probability]

Step 4: Perform the Calculation

Substitute the values identified in Step 1 and Step 2 into the formula:

$P(\text{Spade}) = \frac{13}{52}$

To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is $13$:

$P(\text{Spade}) = \frac{13 \div 13}{52 \div 13}$

$P(\text{Spade}) = \frac{1}{4}$

Final Answer: The probability of drawing a spade is $\frac{1}{4}$.

Given: An event in the context of probability theory that is impossible to occur.

To Find: The numerical probability of such an event and the specific terminology used to describe it.

Step 1: Defining the Probability of an Event
In probability theory, the probability of an event $E$, denoted by $P(E)$, is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes in the sample space $S$.
Mathematically, $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.

Step 2: Analyzing an Impossible Event
An event that cannot happen is defined as an event with zero favorable outcomes. Let $E$ be an impossible event.
Since the event cannot occur, the number of favorable outcomes is $0$.
Let $n(S)$ be the total number of possible outcomes in the sample space, where $n(S) > 0$.
Applying the probability formula:
$P(E) = \frac{0}{n(S)}$
$P(E) = 0$

Step 3: Terminology
In the axiomatic definition of probability, an event that has no possibility of occurring is formally referred to as an impossible event. By definition, the probability of an impossible event is always $0$.

Step 4: Completing the Statement
Based on the deductions above:
1. The probability of an event that cannot happen is $0$.
2. Such an event is called an impossible event.

Final Answer: The probability of an event that cannot happen is 0. Such an event is called an impossible event.

Given: A fair six-faced die is thrown twice. The total number of possible outcomes for a single throw is $6$.

To Find: The probability that the number $5$ will come up at least once in the two throws.

Step 1: Determining the Total Number of Possible Outcomes

When a die is thrown twice, the total number of outcomes is calculated by the product of the outcomes of each throw. Since each throw has $6$ possible outcomes ($1, 2, 3, 4, 5, 6$):

Total outcomes = $6 \times 6 = 36$.

The sample space $S$ is represented as follows:

Step 2: Identifying Favorable Outcomes

Let $E$ be the event that $5$ comes up at least once. This includes outcomes where $5$ appears on the first die, the second die, or both.

Outcomes where $5$ appears on the first die: $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$ [Total: $6$ outcomes].

Outcomes where $5$ appears on the second die: $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)$ [Total: $6$ outcomes].

Note: The outcome $(5,5)$ is common to both lists. To avoid double-counting, we list the unique favorable outcomes:

Favorable outcomes = $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)\}$.

Counting these, we find the number of favorable outcomes $n(E) = 11$.

Step 3: Calculating the Probability

The formula for the probability of an event $E$ is given by:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

[Using the classical definition of probability]

$P(E) = \frac{11}{36}$

Step 4: Verification

Alternatively, we can use the complement rule. Let $E'$ be the event that $5$ never comes up. The outcomes that do not contain a $5$ are those where each die shows $1, 2, 3, 4,$ or $6$ ($5$ possibilities per die).

Number of outcomes without $5 = 5 \times 5 = 25$.

$P(E') = \frac{25}{36}$.

Since $P(E) = 1 - P(E')$, we have $P(E) = 1 - \frac{25}{36} = \frac{36-25}{36} = \frac{11}{36}$.

Final Answer: The probability that 5 will come up at least once is $\frac{11}{36}$.

Given: A set of values representing potential probabilities of an event: (A) $2/3$, (B) $-1.5$, (C) $15\%$, (D) $0.7$.

To Find: Identify which of the given values cannot represent the probability of an event.

Theoretical Background:

In the theory of probability, for any event $E$, the probability $P(E)$ must satisfy the following fundamental axiom:

$0 \leq P(E) \leq 1$

This implies that:

1. The probability of an event cannot be negative ($P(E) \geq 0$).

2. The probability of an event cannot exceed $1$ ($P(E) \leq 1$).

Step 1: Analyzing the given options

We evaluate each option against the condition $0 \leq P(E) \leq 1$.

Step 2: Justification for the invalid value

The value $-1.5$ is less than $0$. Since the probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes, it must always be a non-negative value. Therefore, a negative value is mathematically impossible for a probability.

Conclusion:

Since $-1.5 < 0$, it violates the axiom $P(E) \geq 0$. Thus, it cannot be the probability of an event.

Final Answer: The value that cannot be the probability of an event is -1.5.

Given: A die has six faces marked with the following letters: $A, B, C, D, E, A$.

To Find: The probability of getting the letter $D$ when the die is thrown once.

Visual Representation of the Sample Space:

Step 1: Define the Sample Space ($S$)
The sample space consists of all possible outcomes when the die is thrown. Based on the given faces:
$S = \{A, B, C, D, E, A\}$
The total number of possible outcomes, denoted by $n(S)$, is the count of elements in the set $S$.
$n(S) = 6$ [Since there are 6 faces on the die]

Step 2: Define the Favorable Event ($E$)
Let $E$ be the event of getting the letter $D$.
Looking at the sample space $S = \{A, B, C, D, E, A\}$, we identify the occurrences of $D$.
The favorable outcomes are $\{D\}$.
The number of favorable outcomes, denoted by $n(E)$, is:
$n(E) = 1$

Step 3: Apply the Probability Formula
The theoretical probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.
Formula: $P(E) = \frac{n(E)}{n(S)}$ [Definition of Probability for equally likely outcomes]

Step 4: Calculation
Substitute the values obtained in Step 1 and Step 2 into the formula:
$P(D) = \frac{1}{6}$

Final Answer: The probability of getting D is $\frac{1}{6}$.