Given:

Number of defective pens = $12$

Number of good pens = $132$

To Find:

The probability that a pen taken out at random is a good one.

Step 1: Determine the total number of outcomes.

Let $n(D)$ be the number of defective pens and $n(G)$ be the number of good pens.

Total number of pens in the lot, denoted by $n(S)$, is the sum of defective and good pens.

$n(S) = n(D) + n(G)$

$n(S) = 12 + 132$

$n(S) = 144$

[Since the total number of possible outcomes is the sum of all individual items in the sample space]

Step 2: Define the event and identify favorable outcomes.

Let $E$ be the event of drawing a good pen.

The number of favorable outcomes for event $E$, denoted by $n(E)$, is equal to the number of good pens.

$n(E) = 132$

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.

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

$P(E) = \frac{132}{144}$

Step 4: Simplify the fraction.

To simplify $\frac{132}{144}$, we find the greatest common divisor (GCD) of $132$ and $144$.

Divide both numerator and denominator by their common factors:

Divide by $12$:

$132 \div 12 = 11$

$144 \div 12 = 12$

Therefore, $P(E) = \frac{11}{12}$

Final Answer: The probability that the pen taken out is a good one is $\frac{11}{12}$.

Given:

Total number of ball pens in the lot = $144$

Number of defective pens = $20$

Condition for buying: Nuri buys the pen if it is good; she does not buy the pen if it is defective.

To Find:

The probability that Nuri will not buy the pen.

Step 1: Defining the Sample Space

Let $S$ be the sample space representing the total number of pens available in the lot. The total number of outcomes is given by the total number of pens.

$n(S) = 144$

Step 2: Defining the Event

Let $E$ be the event that Nuri will not buy the pen. According to the problem statement, Nuri will not buy the pen if it is defective.

Therefore, the number of favorable outcomes for event $E$ is equal to the number of defective pens.

$n(E) = 20$

Step 3: Applying 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 in the sample space.

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

Substituting the known values:

$P(E) = \frac{20}{144}$

Step 4: Simplifying the Fraction

To simplify $\frac{20}{144}$, we find the greatest common divisor (GCD) of $20$ and $144$.

Prime factorization of $20$: $2^2 \times 5$

Prime factorization of $144$: $2^4 \times 3^2$

The common factor is $2^2 = 4$.

Dividing both the numerator and the denominator by $4$:

$P(E) = \frac{20 \div 4}{144 \div 4}$

$P(E) = \frac{5}{36}$

Final Answer: The probability that Nuri will not buy the pen is $\frac{5}{36}$.

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 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 a number greater than $2$.

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

Step 2: Define the Event
Let $E$ be the event of getting a number greater than $2$.
The numbers in the sample space that are greater than $2$ are $\{3, 4, 5, 6, 7, 8\}$.
Therefore, $E = \{3, 4, 5, 6, 7, 8\}$.

Step 3: Count the Favorable Outcomes
The number of favorable outcomes, denoted by $n(E)$, is the count of elements in set $E$.
$n(E) = 6$

Step 4: 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 5: Calculate the Probability
Substitute the values obtained in Step 1 and Step 3 into the formula:
$P(E) = \frac{6}{8}$
To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is $2$:
$P(E) = \frac{6 \div 2}{8 \div 2} = \frac{3}{4}$

Final Answer: The probability that the arrow will point at a number greater than 2 is $\frac{3}{4}$ (or $0.75$).

Given: An experiment where a baby is born, and the outcome is either a boy or a girl.

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 $n$ possible 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 baby is a boy.
$E_2$: The baby is a girl.
Thus, $S = \{ \text{Boy, Girl} \}$. The total number of possible outcomes is $n = 2$.

Step 2: Analyzing the Probability of Each Outcome
In biological terms, the sex of a baby is determined by the combination of sex chromosomes (XX for female, XY for male). During fertilization, there is an approximately equal chance of a sperm carrying an X chromosome or a Y chromosome fertilizing the egg.
Let $P(E_1)$ be the probability that the baby is a boy.
Let $P(E_2)$ be the probability that the baby is a girl.

Step 3: Evaluating the Likelihood
Based on biological principles and statistical data collected over large populations, the ratio of male births to female births is approximately $1:1$.
Therefore:
$P(E_1) \approx 0.5$
$P(E_2) \approx 0.5$
Since $P(E_1) = P(E_2) = \frac{1}{2}$, the outcomes are considered to have an equal chance of occurrence.

Step 4: Conclusion
Because the probability of the baby being a boy is equal to the probability of the baby being a girl, the outcomes are equally likely.

Final Answer: Yes, the outcomes are equally likely because, biologically, the chance of a baby being a boy or a girl is approximately equal ($50\%$ each).

Given:

Total number of bulbs in the lot ($n(S)$) = $20$

Number of defective bulbs ($n(E)$) = $4$

To find:

The probability that the bulb drawn at random is defective ($P(E)$).

Step 1: Defining the Sample Space

The sample space $S$ consists of all the bulbs in the lot. Since there are $20$ bulbs in total, the total number of possible outcomes is:

$n(S) = 20$

Step 2: Defining the Event

Let $E$ be the event of drawing a defective bulb. The number of favorable outcomes for this event is the number of defective bulbs present in the lot:

$n(E) = 4$

Step 3: Applying the Probability Formula

The theoretical 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:

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

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

Step 4: Calculation

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

$P(E) = \frac{4}{20}$

To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is $4$:

$P(E) = \frac{4 \div 4}{20 \div 4}$

$P(E) = \frac{1}{5}$

Converting the fraction to decimal form (optional but precise):

$P(E) = 0.2$

Final Answer: The probability that the bulb drawn is defective is $\frac{1}{5}$ or $0.2$.

Given: A trial is conducted to answer a true-false question. The possible outcomes are that the answer is either 'right' or 'wrong'.

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

Definition: Outcomes of an experiment are said to be equally likely if each outcome has the same probability of occurring. In a sample space $S$, if $P(A) = P(B) = \dots = P(n)$, then the outcomes are equally likely.

Step 1: Identifying the Sample Space
Let the possible outcomes of the experiment be represented by the set $S$.
$S = \{ \text{Right}, \text{Wrong} \}$
The total number of possible outcomes is $n(S) = 2$.

Step 2: Analyzing the Nature of the Experiment
In a true-false question, the correctness of the answer depends on the knowledge or the choice of the person answering the question. Unlike a fair coin toss where the physical properties of the coin ensure a $50\%$ chance for heads or tails, the probability of answering a true-false question correctly is dependent on external factors:

• If the person knows the subject matter, the probability of answering 'Right' is significantly higher than 'Wrong'.
• If the person is guessing, the probability might be $0.5$ for each.
• If the person is uninformed, the probability of answering 'Wrong' might be higher.

Step 3: Evaluating the Condition for Equally Likely Outcomes
For the outcomes to be equally likely, the probability of getting the answer 'Right' ($P(R)$) must equal the probability of getting the answer 'Wrong' ($P(W)$).
$P(R) = \frac{\text{Number of favorable outcomes for Right}}{\text{Total number of outcomes}}$
$P(W) = \frac{\text{Number of favorable outcomes for Wrong}}{\text{Total number of outcomes}}$

Since the probability of answering correctly is not inherently fixed at $0.5$ for every individual or every question (as it is not a random physical process like tossing a balanced die), we cannot assume $P(R) = P(W) = 0.5$.

Step 4: Conclusion
Because the likelihood of the answer being 'Right' or 'Wrong' depends on the knowledge of the person answering the question and is not determined by a uniform random process, the outcomes are not necessarily equally likely.

Final Answer: The outcomes are not equally likely because the probability of answering correctly depends on the knowledge of the person attempting the question, and it is not a random event with a fixed probability of $0.5$ for each outcome.

Given:

The number of red marbles in the box = $5$

The number of white marbles in the box = $8$

The number of green marbles in the box = $4$

To find:

The probability that the marble taken out at random is not green.

Step 1: Calculate the total number of outcomes.

The total number of marbles in the box represents the total number of possible outcomes ($n(S)$).

$n(S) = \text{Number of red marbles} + \text{Number of white marbles} + \text{Number of green marbles}$

$n(S) = 5 + 8 + 4$

$n(S) = 17$

Step 2: Define the event and calculate the number of favorable outcomes.

Let $E$ be the event that the marble taken out is not green.

A marble is "not green" if it is either red or white.

Number of favorable outcomes ($n(E)$) = $\text{Number of red marbles} + \text{Number of white marbles}$

$n(E) = 5 + 8$

$n(E) = 13$

Step 3: Apply the probability formula.

The probability of an event $E$ is given by the ratio of the number of favorable outcomes to the total number of possible outcomes:

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

$P(\text{not green}) = \frac{13}{17}$

Alternative Method (Using Complementary Events):

Let $G$ be the event that the marble taken out is green.

$n(G) = 4$

$P(G) = \frac{n(G)}{n(S)} = \frac{4}{17}$

The event "not green" is the complement of the event "green", denoted as $P(\overline{G})$.

$P(\overline{G}) = 1 - P(G)$ [Since the sum of probabilities of complementary events is 1]

$P(\overline{G}) = 1 - \frac{4}{17}$

$P(\overline{G}) = \frac{17 - 4}{17}$

$P(\overline{G}) = \frac{13}{17}$

Final Answer: The probability that the marble taken out will not be green is $\frac{13}{17}$.

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 rectangular region with dimensions length $l = 3$ m and breadth $b = 2$ m. A circular region is inscribed within this rectangle with a diameter $d = 1$ m.

To Find:

The probability that a die dropped at random on the rectangular region will land inside the circle.

Visual Representation:

Step 1: Calculate the Area of the Rectangular Region

The formula for the area of a rectangle is $Area = length \times breadth$.

Given $l = 3$ m and $b = 2$ m:

$Area_{rectangle} = 3 \text{ m} \times 2 \text{ m} = 6 \text{ m}^2$.

Step 2: Calculate the Area of the Circular Region

The formula for the area of a circle is $Area = \pi r^2$, where $r$ is the radius.

Given the diameter $d = 1$ m, the radius $r$ is calculated as:

$r = \frac{d}{2} = \frac{1}{2} = 0.5 \text{ m}$.

Now, calculate the area:

$Area_{circle} = \pi \times (0.5 \text{ m})^2$

$Area_{circle} = \pi \times 0.25 \text{ m}^2 = \frac{\pi}{4} \text{ m}^2$.

Step 3: Apply the Probability Formula for Geometric Regions

The probability $P(E)$ of an event occurring in a geometric region is defined as:

$P(E) = \frac{\text{Favorable Area}}{\text{Total Area}}$

[Since the die is dropped at random, the probability is proportional to the area of the target region relative to the total area].

Step 4: Perform the Final Calculation

$P(\text{landing inside circle}) = \frac{Area_{circle}}{Area_{rectangle}}$

$P = \frac{\frac{\pi}{4}}{6}$

$P = \frac{\pi}{4 \times 6}$

$P = \frac{\pi}{24}$

Final Answer: The probability that the die will land inside the circle is $\frac{\pi}{24}$.

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:

• Total number of ball pens in the lot ($n(S)$) = $144$
• Number of defective pens = $20$

To Find:

The probability that Nuri will buy the pen. Nuri buys the pen if and only if it is a good pen.

Step 1: Determine the number of good pens.

Let $N_{total}$ be the total number of pens and $N_{defective}$ be the number of defective pens.

The number of good pens ($N_{good}$) is calculated as:

$N_{good} = N_{total} - N_{defective}$

$N_{good} = 144 - 20$

$N_{good} = 124$

[Since the total lot consists only of good and defective pens, subtracting the defective count from the total yields the count of good pens.]

Step 2: Define the event and the probability formula.

Let $E$ be the event that Nuri buys the pen. Nuri buys the pen if it is good.

The number of favorable outcomes for event $E$ is the number of good pens, which is $124$.

The total number of possible outcomes is the total number of pens, which is $144$.

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

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

Step 3: Calculate the probability.

$P(\text{She will buy it}) = \frac{124}{144}$

To simplify the fraction, we find the greatest common divisor (GCD) of $124$ and $144$.

Divide both numerator and denominator by $4$:

$124 \div 4 = 31$

$144 \div 4 = 36$

$P(E) = \frac{31}{36}$

[Since $31$ is a prime number and does not divide $36$, the fraction is in its simplest form.]

Final Answer: The probability that she will buy the pen is $\frac{31}{36}$.

Given: A football game begins with a coin toss to decide which team gets the ball first. The coin has two distinct faces: Heads (H) and Tails (T).

To Find: The logical and mathematical justification for why a coin toss is considered a "fair" method of decision-making.

Visual Representation of the Sample Space:

Step 1: Defining the Sample Space

When a fair coin is tossed, there are only two possible outcomes: Heads ($H$) and Tails ($T$). Therefore, the sample space $S$ is defined as:
$S = \{H, T\}$
The total number of possible outcomes, denoted by $n(S)$, is $2$.

Step 2: Calculating Probabilities of Individual Events

Let $E_1$ be the event of getting Heads and $E_2$ be the event of getting Tails.
The probability of an event $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For Heads ($E_1$):
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{2} = 0.5$

For Tails ($E_2$):
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{2} = 0.5$

Step 3: Justification of Fairness

A process is considered "fair" in probability theory if all possible outcomes are equally likely.
Since $P(E_1) = P(E_2) = 0.5$, neither outcome has a mathematical advantage over the other. Because the probability of obtaining Heads is exactly equal to the probability of obtaining Tails, the outcome of the toss is entirely dependent on chance and cannot be predicted or manipulated by either team.

Step 4: Conclusion

Because the two outcomes are mutually exclusive and collectively exhaustive with equal probabilities, the coin toss provides an unbiased mechanism to decide which team gains possession of the ball. This eliminates any subjective bias or preference, ensuring that both teams have an equal opportunity ($50\%$) to win the toss.

Final Answer: Tossing a coin is considered a fair way to decide because the two possible outcomes, Heads and Tails, are equally likely, each having a probability of $\frac{1}{2}$. This ensures that neither team has an advantage, making the decision process unbiased and purely based on chance.

Given: A game of chance involves a spinner with 8 equally likely outcomes, represented by the set of numbers $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$.

To Find: The probability that the arrow points at a number less than 9.

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

Step 2: Define the Event
Let $E$ be the event that the arrow points at a number less than 9.
By examining the set $S$, we identify the numbers that satisfy the condition $x < 9$:
$E = \{1, 2, 3, 4, 5, 6, 7, 8\}$
The number of favorable outcomes, denoted by $n(E)$, is 8.

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 for equally likely outcomes]

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

Justification: Since every number on the spinner is less than 9, the event $E$ is a sure event (or certain event). The probability of a sure event is always 1.

Final Answer: The probability that the arrow will point at a number less than 9 is 1.

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

To Find: The probability of drawing a king of red colour.

Step 1: Determine the Total Number of Possible Outcomes

In a standard deck of playing cards, the total number of cards is $52$. Since one card is drawn at random, the total number of possible outcomes ($n(S)$) is:

Step 2: Identify the Favorable Outcomes

We are looking for the probability of drawing a "king of red colour".

A standard deck consists of four suits: Hearts, Diamonds, Spades, and Clubs.

• Hearts and Diamonds are red suits.
• Spades and Clubs are black suits.

Each suit contains exactly one king. Therefore:

• The king of Hearts is a red king.
• The king of Diamonds is a red king.

Let $E$ be the event of drawing a king of red colour. The number of favorable outcomes ($n(E)$) is:

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:

Step 4: Perform the Calculation

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

Now, simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is $2$:

Final Answer: The probability of getting a king of red colour is $\mathbf{\frac{1}{26}}$.

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 box contains 90 discs, numbered from 1 to 90. One disc is drawn at random.

To Find: The probability that the drawn disc bears a number divisible by 5.

Step 1: Determine the Total Number of Possible Outcomes

Since the discs are numbered from 1 to 90, the total number of discs in the box is 90. Therefore, the total number of possible outcomes ($n(S)$) is:

$n(S) = 90$

Step 2: Identify the Favorable Outcomes

Let $E$ be the event of drawing a disc with a number divisible by 5. The numbers between 1 and 90 that are divisible by 5 form an arithmetic progression (AP):

$5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90$

To find the count of these numbers, we use the formula for the $n^{th}$ term of an arithmetic progression: $a_n = a + (n - 1)d$.

Here, the first term ($a$) = 5, the common difference ($d$) = 5, and the last term ($a_n$) = 90.

$90 = 5 + (n - 1)5$

[Subtract 5 from both sides]

$85 = (n - 1)5$

[Divide both sides by 5]

$17 = n - 1$

[Add 1 to both sides]

$n = 18$

Thus, the number of favorable outcomes ($n(E)$) is 18.

Step 3: Calculate the Probability

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

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

[Substituting the values obtained in Step 1 and Step 2]

$P(E) = \frac{18}{90}$

Step 4: Simplify the Fraction

[Divide both numerator and denominator by their greatest common divisor, which is 18]

$P(E) = \frac{18 \div 18}{90 \div 18}$

$P(E) = \frac{1}{5}$

Final Answer: The probability that the drawn disc bears a number divisible by 5 is $\frac{1}{5}$ or 0.2.

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 with a set of all possible elementary events.

To Find: The sum of the probabilities of all the elementary events of the experiment.

Step 1: Defining Elementary Events
An elementary event is defined as an event which has only one possible outcome of an experiment. Let the set of all possible elementary events of a random experiment be denoted by $E_1, E_2, E_3, \dots, E_n$.

Step 2: Applying the Axioms of Probability
According to the fundamental axioms of probability theory, for any random experiment, the sum of the probabilities of all the elementary events must equal the probability of the sure event (the sample space itself).

Step 3: Mathematical Representation
Let $P(E_i)$ represent the probability of the occurrence of the $i$-th elementary event. The sum of all such probabilities is given by:
$\sum_{i=1}^{n} P(E_i) = P(E_1) + P(E_2) + P(E_3) + \dots + P(E_n)$

Step 4: Logical Deduction
Since the set $\{E_1, E_2, \dots, E_n\}$ encompasses all possible outcomes of the experiment, the union of these events constitutes the sample space $S$. The probability of the sample space $P(S)$ is always $1$, as it represents a sure or certain event.

Therefore, $\sum_{i=1}^{n} P(E_i) = 1$.

Conclusion:
The sum of the probabilities of all the elementary events of an experiment is always equal to $1$.

Final Answer: 1

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 an odd number.

Step 1: Defining the Sample Space
Let $S$ be the sample space representing all possible outcomes when a die is thrown. Since a standard die has six faces numbered 1 through 6:
$S = \{1, 2, 3, 4, 5, 6\}$
The total number of possible outcomes, denoted by $n(S)$, is:
$n(S) = 6$

Step 2: Defining the Event
Let $E$ be the event of getting an odd number. An odd number is an integer that is not divisible by 2. From the sample space $S$, we identify the odd numbers:
$E = \{1, 3, 5\}$
The number of favorable outcomes, denoted by $n(E)$, is:
$n(E) = 3$

Step 3: Applying 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 in the sample space:
$P(E) = \frac{n(E)}{n(S)}$
[Using the classical definition of probability for equally likely outcomes]

Step 4: Calculation
Substitute the values obtained in Step 1 and Step 2 into the formula:
$P(E) = \frac{3}{6}$
To simplify the fraction, divide 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 an odd number is $\frac{1}{2}$ (or 0.5).