Probability — Notes

14.1 Probability — A Theoretical Approach

Probability is the measure of the likelihood that an event will occur. It quantifies uncertainty. In mathematics, we often deal with theoretical probability, which is based on logical reasoning and assumptions, rather than experimental results.

Key Concepts in Probability

• Experiment: An action or process that produces observable outcomes.

• Example: Tossing a coin, rolling a die, drawing a card.

• Outcome: A single possible result of an experiment.

• Example: For a coin toss, outcomes are Head (H) or Tail (T). For a die roll, outcomes are 1, 2, 3, 4, 5, or 6.

• Sample Space (S): The set of all possible outcomes of an experiment.

• Example: For tossing a coin, $S=\{H,T\}$. For rolling a die, $S=\{1,2,3,4,5,6\}$.

• Event (E): A collection of outcomes of an experiment. An event can be a single outcome or multiple outcomes.

• Example: Getting a head when tossing a coin ($E=\{H\}$). Getting an even number when rolling a die ($E=\{2,4,6\}$).

• Equally Likely Outcomes: Outcomes that have the same chance of occurring. This is a crucial assumption for theoretical probability.

• Why it matters: If outcomes are not equally likely (e.g., a biased coin), the standard probability formula doesn't apply directly, and we'd need experimental data or more complex methods.

• Example: A fair coin toss has equally likely outcomes (Head and Tail). A fair die roll has equally likely outcomes (1, 2, 3, 4, 5, 6).

• Counter-example: A bag with 4 red balls and 1 blue ball. Drawing a red ball is more likely than drawing a blue ball, so these outcomes are not equally likely.

• Theoretical Probability (Classical Probability): The probability of an event E is calculated as the ratio of the number of outcomes favourable to E to the total number of possible outcomes in the sample space.

This formula is valid only when the outcomes are equally likely.

Example 1: Coin Toss

Given: A fair coin is tossed once. To find: The probability of getting a Head and the probability of getting a Tail. Solution:

• Experiment: Tossing a coin once.
• Sample Space (S): $\{H,T\}$. Total number of possible outcomes = 2.
• Event E (Getting a Head): $\{H\}$. Number of outcomes favourable to E = 1.

• Event F (Getting a Tail): $\{T\}$. Number of outcomes favourable to F = 1.

Remember

For a fair coin, $P(\text{Head})=P(\text{Tail})=\frac{1}{2}$.

Exam-smart

Case studies might describe a coin and ask for probabilities. Always check if it's "fair" or "biased". If "fair", use the formula. If "biased", the problem must give you the specific probabilities.

Example 2: Coloured Balls in a Bag

Given: A bag contains a red ball, a blue ball, and a yellow ball, all of the same size. One ball is drawn without looking. To find: The probability of drawing a yellow ball, a red ball, and a blue ball. Solution:

• Experiment: Drawing one ball from the bag.
• Sample Space (S): {Red, Blue, Yellow}. Total number of possible outcomes = 3 (since they are of the same size, each is equally likely).
• Event Y (Yellow ball): {Yellow}. Number of favourable outcomes = 1.

• Event R (Red ball): {Red}. Number of favourable outcomes = 1.

• Event B (Blue ball): {Blue}. Number of favourable outcomes = 1.

Remember

The assumption of "same size" or "well-shuffled" is critical for ensuring equally likely outcomes.

Exam-smart

Word problems often involve selecting items (balls, cards, etc.). Carefully count the total items and the specific items requested to find favourable outcomes.

Example 3: Rolling a Die

Given: A fair die is thrown once. To find: (i) Probability of getting a number greater than 4. (ii) Probability of getting a number less than or equal to 4.

Solution:

• Experiment: Rolling a fair die once.

• Sample Space (S): $\{1,2,3,4,5,6\}$. Total number of possible outcomes = 6.

• (i) Event E (Getting a number > 4):

• Outcomes favourable to E: $\{5,6\}$. Number of favourable outcomes = 2.

• (ii) Event F (Getting a number ≤ 4):
• Outcomes favourable to F: $\{1,2,3,4\}$. Number of favourable outcomes = 4.

Remember

Always simplify your probability fractions.

Exam-smart

Questions might ask for ranges of numbers (e.g., "between 2 and 6", "greater than 4"). List these outcomes carefully.

Types of Events

• Elementary Event: An event that has only one outcome.

• Example: Getting a '3' when rolling a die. In Example 1, 'Head' and 'Tail' are elementary events. In Example 2, 'Yellow', 'Red', and 'Blue' are elementary events.

• Sure Event (Certain Event): An event that is sure to happen. Its probability is 1.

• Example: Getting a number less than 7 when rolling a die (since all outcomes are 1, 2, 3, 4, 5, 6, which are all less than 7).

• Calculation: Favourable outcomes = 6, Total outcomes = 6. $P(\text{Sure Event})=\frac{6}{6}=1$.

• Impossible Event: An event that cannot happen. Its probability is 0.

• Example: Getting an 8 when rolling a standard die.

• Calculation: Favourable outcomes = 0, Total outcomes = 6. $P(\text{Impossible Event})=\frac{0}{6}=0$.

• Complementary Events: Two events are complementary if one event happening means the other cannot happen, and vice versa. If E is an event, its complement is denoted by $E^{\mathrm{\prime }}$ or $\bar{E}$ (read as "not E").

• Property: The sum of the probabilities of an event and its complement is always 1.

• This implies:

• Why it matters: This is very useful for calculating probabilities of "not happening" events indirectly. If it's hard to count favourable outcomes for an event, it's often easier to count favourable outcomes for its complement and subtract from 1.

Example 4: Playing Cards

Given: One card is drawn from a well-shuffled deck of 52 cards. To find: The probability that the card is (i) an ace, (ii) not an ace.

Solution:

• Experiment: Drawing one card from a deck of 52 cards.

• Sample Space (S): 52 cards. Total number of possible outcomes = 52.

• (i) Event E (Card is an ace):

• There are 4 aces in a deck (Ace of Spades, Hearts, Diamonds, Clubs).

• Number of favourable outcomes = 4.

• (ii) Event F (Card is not an ace):
• This is the complement of event E ($E^{\mathrm{\prime }}$).
• Number of outcomes not favourable to E = Total cards - Number of aces = $52-4=48$.
• So, number of favourable outcomes for F = 48.

• Alternatively, using the complement rule:

Remember

Always remember that $0\le P(E)\le 1$. A probability cannot be negative or greater than 1.

Exam-smart

Cards problems are common. Know the composition of a standard deck: 52 cards, 4 suits (Spades ♠, Hearts ♥, Diamonds ♦, Clubs ♣), 2 colours (Red: Hearts & Diamonds; Black: Spades & Clubs), 13 ranks (Ace, 2-10, Jack, Queen, King), and 4 face cards (J, Q, K) in each suit.

Example 5: Sangeeta vs. Reshma Tennis Match

Given: Probability of Sangeeta winning a match is $P(S)=0.62$. To find: Probability of Reshma winning the match. Solution:

• Experiment: Playing a tennis match between Sangeeta and Reshma.
• Possible Outcomes: Sangeeta wins (S) or Reshma wins (R). These are the only two possibilities and they are mutually exclusive. Thus, they are complementary events.
• Sample Space: $\{S,R\}$.
• We know $P(S)+P(R)=1$ (since they are complementary events).

Remember

In a situation with only two possible outcomes, they are complementary.

Exam-smart

These problems test the understanding of complementary events in real-world contexts like games or outcomes.

Example 6: Birthdays

Given: Two friends, Savita and Hamida. To find: Probability of them having (i) different birthdays, (ii) the same birthday (ignoring leap years). Solution:

• Experiment: Considering the birthdays of Savita and Hamida.

• Assumption: 365 days in a year, each equally likely for a birthday.

• Total Possible Outcomes: Hamida's birthday can be any of the 365 days. So, total outcomes = 365.

• (i) Event E (Different birthdays):

• Savita's birthday can be any day. For Hamida to have a different birthday, her birthday must not be on Savita's birthday.

• Number of favourable outcomes for Hamida's birthday = $365-1=364$.

• (ii) Event F (Same birthday):
• This is the complement of Event E (different birthdays).

Remember

In problems involving days or selections, carefully identify the total possibilities and the favourable outcomes.

Exam-smart

Birthday problems often highlight the concept of complementary events and large sample spaces.

Example 7: Class Representative Selection

Given: A Class X has 40 students (25 girls, 15 boys). One student is selected as class representative by drawing a card with names. To find: Probability that the selected student is (i) a girl, (ii) a boy.

Solution:

• Experiment: Selecting one student's name card from the bag.

• Sample Space (S): 40 students' names. Total number of possible outcomes = 40.

• (i) Event G (Selecting a girl):

• Number of girls = 25. These are the favourable outcomes.

• (ii) Event B (Selecting a boy):
• Number of boys = 15. These are the favourable outcomes.

• Alternatively, using complement: $P(B)=1-P(G)=1-\frac{5}{8}=\frac{3}{8}$.

Remember

Ensure the total number of outcomes matches the total number of items from which selection is made.

Exam-smart

This is a typical problem involving proportions in a population. Calculate probabilities based on given counts.

Example 8: Marbles in a Box

Given: A box contains 3 blue, 2 white, and 4 red marbles. One marble is drawn at random. To find: Probability of drawing (i) a white marble, (ii) a blue marble, (iii) a red marble.

Solution:

• Experiment: Drawing one marble at random from the box.

• Total number of marbles: $3(\text{blue})+2(\text{white})+4(\text{red})=9$.

• Sample Space (S): 9 marbles. Total number of possible outcomes = 9.

• (i) Event W (White marble):

• Number of white marbles = 2.

• (ii) Event B (Blue marble):
• Number of blue marbles = 3.

• (iii) Event R (Red marble):
• Number of red marbles = 4.

Remember

Check that the sum of probabilities of all elementary events equals 1: $P(W)+P(B)+P(R)=\frac{2}{9}+\frac{3}{9}+\frac{4}{9}=\frac{9}{9}=1$.

Exam-smart

Similar to ball problems, these require careful counting of items of different types.

Example 9: Tossing Two Coins

Given: Two different coins are tossed simultaneously. To find: Probability of getting at least one head.

Solution:

• Experiment: Tossing two coins simultaneously.

• Sample Space (S): Let H be Head, T be Tail. The possible outcomes are:

• Coin 1 = H, Coin 2 = H $\to$ (H, H)

• Coin 1 = H, Coin 2 = T $\to$ (H, T)

• Coin 1 = T, Coin 2 = H $\to$ (T, H)

• Coin 1 = T, Coin 2 = T $\to$ (T, T) So, $S=\{(H,H),(H,T),(T,H),(T,T)\}$. Total number of possible outcomes = 4.

• Event E (Getting at least one head): This means one head OR two heads.

• Outcomes favourable to E: $\{(H,H),(H,T),(T,H)\}$.

• Number of favourable outcomes = 3.

• Alternative Method (Using Complement):
• The complement event E' is "getting no heads", which means getting all tails.
• Outcome favourable to E': $\{(T,T)\}$. Number of favourable outcomes = 1.

• Using the complement rule:

Remember

When dealing with multiple coins or dice, systematically list all possible outcomes. The total number of outcomes for $n$ coins is $2^n$. For $n$ dice, it's $6^n$.

Exam-smart

"At least one" problems are usually best solved using the complement rule: $P(\text{at least one})=1-P(\text{none})$.

Introduction to Continuous Probability (Brief Mention)

The definition of theoretical probability $P(E)=\frac{\text{Favourable Outcomes}}{\text{Total Outcomes}}$ works well for experiments with a finite number of outcomes. However, in situations where outcomes can be any value within a range (e.g., time, length, area), there are infinitely many possibilities.

• Example 10 (Musical Chair): If music stops any time between 0 and 2 minutes, the sample space is the interval $[0,2]$. The probability of stopping in the first half-minute ($[0,0.5]$) is the ratio of the length of the favourable interval to the length of the total interval: $\frac{0.5}{2}=\frac{1}{4}$.

• Example 11 (Helicopter Crash): If a helicopter can crash anywhere in a rectangular region, and we want the probability of it crashing in a specific part (like a lake), we use the ratio of areas.

• Area of lake = $7.5{\text{km}}^2$

• Area of region = $40.5{\text{km}}^2$

• $P(\text{Crash in lake})=\frac{7.5}{40.5}=\frac{75}{405}=\frac{5}{27}$.

Note: These examples (10 and 11) are for conceptual understanding and are usually not part of the standard examination syllabus for this chapter at the Class 9-10 level. The focus remains on discrete probability.

Remember

For continuous outcomes, probability is calculated using measures like length, area, or volume, not by counting discrete outcomes.

Exam-smart

While these continuous probability examples are in the textbook, expect questions based on discrete outcomes for board exams.

Example 12: Shirt Defects

Given: A carton has 100 shirts: 88 good, 8 with minor defects, 4 with major defects. Jimmy accepts good shirts; Sujatha rejects shirts with major defects. One shirt is drawn at random. To find: Probability that the shirt is (i) acceptable to Jimmy, (ii) acceptable to Sujatha.

Solution:

• Experiment: Drawing one shirt at random from the carton.

• Sample Space (S): 100 shirts. Total number of possible outcomes = 100.

• (i) Event J (Acceptable to Jimmy):

• Jimmy accepts only good shirts. Number of good shirts = 88.

• (ii) Event Su (Acceptable to Sujatha):
• Sujatha rejects shirts with major defects. This means she accepts shirts that are good or have minor defects.
• Number of shirts acceptable to Sujatha = Number of good shirts + Number of shirts with minor defects
• Number of favourable outcomes = $88+8=96$.

Remember

Read the acceptance/rejection criteria carefully to determine the favourable outcomes.

Exam-smart

These problems involve categorizing items based on given criteria and then calculating probabilities.

Example 13: Tossing Two Dice

Given: Two dice, one blue and one grey, are thrown simultaneously. To find: Probability that the sum of the numbers appearing on the top of the dice is (i) 8, (ii) 13, (iii) less than or equal to 12.

Solution:

• Experiment: Throwing two dice simultaneously.
• Sample Space (S): Each die has 6 outcomes. The total number of outcomes is $6\times 6=36$. We can represent them as ordered pairs $(x,y)$, where $x$ is the outcome on the blue die and $y$ is the outcome on the grey die.

Total outcomes = 36.

• (i) Event E (Sum of numbers is 8):
• Favourable outcomes: $(2,6),(3,5),(4,4),(5,3),(6,2)$.
• Number of favourable outcomes = 5.

• (ii) Event F (Sum of numbers is 13):
• The maximum sum possible is $6+6=12$. There is no outcome where the sum is 13.
• Number of favourable outcomes = 0.

• This is an impossible event.

• (iii) Event G (Sum of numbers is less than or equal to 12):

• All possible outcomes result in a sum less than or equal to 12 (maximum sum is 12).

• Number of favourable outcomes = 36.

• This is a sure event.

Remember

For two dice, the total number of outcomes is $6\times 6=36$. For $n$ dice, it's $6^n$.

Exam-smart

Table-based representation is key for dice problems. Sums often form a pattern (e.g., sum 7 is most likely). Be careful with sums that are impossible or certain.

Probability Range and Summary of Properties

• Range of Probability: The probability of any event E, denoted by $P(E)$, is always between 0 and 1, inclusive.

• Summary Table of Properties:

Remember

These properties are fundamental and are often tested in "Fill in the blanks" or "True/False" questions.

Exam-smart

Questions 1 and 2 in Exercise 14.1 directly test these summary properties. Ensure you can state them clearly.

Exercise 14.1 Problems and Solutions

1. Complete the following statements: (i) Probability of an event E + Probability of the event ‘not E’ = 1. (ii) The probability of an event that cannot happen is 0. Such an event is called impossible event. (iii) The probability of an event that is certain to happen is 1. Such an event is called sure event (or certain event). (iv) The sum of the probabilities of all the elementary events of an experiment is 1. (v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

2. Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. Not equally likely. (Depends on car condition, driver's skill, etc.) (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. Not equally likely. (Depends on player's skill.) (iii) A trial is made to answer a true-false question. The answer is right or wrong. Equally likely (assuming the question has a definitive answer and the guess is random). (iv) A baby is born. It is a boy or a girl. Equally likely (biologically, the chances are very close to 50:50, and for probability problems, this is assumed).

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? Tossing a coin is considered fair because a fair coin has two outcomes, Head and Tail, which are equally likely ($P(\text{Head})=P(\text{Tail})=1\mathrm{/}2$). This ensures that each team has an equal chance of winning the toss, without any bias.

4. Which of the following cannot be the probability of an event? (A) $\frac{2}{3}$ (Possible, as $0\le \frac{2}{3}\le 1$) (B) –1.5 (Not possible, as probability cannot be negative) (C) 15% (Possible, as $15\mathrm{\%}=0.15$, and $0\le 0.15\le 1$) (D) 0.7 (Possible, as $0\le 0.7\le 1$) Answer: (B) –1.5

5. If P(E) = 0.05, what is the probability of ‘not E’? Using the complement rule, $P(E^{\mathrm{\prime }})=1-P(E)$. $P(\text{not E})=1-0.05=0.95$. Answer: 0.95

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?

• Sample Space: Candies in the bag (all lemon flavoured).
• Total Outcomes: Let's assume there are $n$ candies.
• (i) Event (Orange candy): There are 0 orange candies. Favourable outcomes = 0.

(This is an impossible event).

• (ii) Event (Lemon candy): All $n$ candies are lemon flavoured. Favourable outcomes = $n$.

(This is a sure event). Answer: (i) 0, (ii) 1

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

• Let E be the event that 2 students do not have the same birthday. We are given $P(E)=0.992$.
• Let E' be the event that the 2 students have the same birthday.
• Since E and E' are complementary events:

Answer: 0.008

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

• Total balls: $3+5=8$.
• (i) Event (Red): Favourable outcomes = 3.

• (ii) Event (Not Red): This means drawing a black ball. Favourable outcomes = 5.

(Alternatively: $P(\text{Not Red})=1-P(\text{Red})=1-\frac{3}{8}=\frac{5}{8}$) Answer: (i) $\frac{3}{8}$, (ii) $\frac{5}{8}$

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

• Total marbles: $5+8+4=17$.
• (i) Event (Red): Favourable outcomes = 5.

• (ii) Event (White): Favourable outcomes = 8.

• (iii) Event (Not Green): This means Red or White. Favourable outcomes = $5+8=13$.

(Alternatively: $P(\text{Not Green})=1-P(\text{Green})=1-\frac{4}{17}=\frac{13}{17}$) Answer: (i) $\frac{5}{17}$, (ii) $\frac{8}{17}$, (iii) $\frac{13}{17}$

10. A piggy bank contains hundred 50p coins, fifty 1 coins, twenty 2 coins and ten 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a 5 coin?

• Total number of coins: $100+50+20+10=180$.
• (i) Event (50 p coin): Favourable outcomes = 100.

• (ii) Event (Not a 5 coin): Favourable outcomes = Total coins - Number of 5 coins $=180-10=170$.

(Alternatively: $P(\text{Not ‘ 5})=1-P(\text{‘ 5})=1-\frac{10}{180}=1-\frac{1}{18}=\frac{17}{18}$) Answer: (i) $\frac{5}{9}$, (ii) $\frac{17}{18}$

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

• Total fish: $5+8=13$.
• Event (Male fish): Favourable outcomes = 5.

Answer: $\frac{5}{13}$

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

• Sample Space (S): $\{1,2,3,4,5,6,7,8\}$. Total outcomes = 8.
• (i) Event (Points at 8): Favourable outcome = {8}. Number = 1.

• (ii) Event (Points at an odd number): Favourable outcomes = {1, 3, 5, 7}. Number = 4.

• (iii) Event (Points at a number > 2): Favourable outcomes = {3, 4, 5, 6, 7, 8}. Number = 6.

• (iv) Event (Points at a number < 9): All numbers in the sample space (1 to 8) are less than 9. Favourable outcomes = 8.

(Sure event) Answer: (i) $\frac{1}{8}$, (ii) $\frac{1}{2}$, (iii) $\frac{3}{4}$, (iv) 1

13. A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

• Sample Space (S): $\{1,2,3,4,5,6\}$. Total outcomes = 6.
• (i) Event (Prime number): Prime numbers in S are {2, 3, 5}. Number = 3.

• (ii) Event (Number lying between 2 and 6): This means numbers strictly greater than 2 and strictly less than 6. Favourable outcomes = {3, 4, 5}. Number = 3.

• (iii) Event (Odd number): Favourable outcomes = {1, 3, 5}. Number = 3.

Answer: (i) $\frac{1}{2}$, (ii) $\frac{1}{2}$, (iii) $\frac{1}{2}$

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds

• Total cards: 52.
• (i) Event (King of red colour): There are two red kings (King of Hearts, King of Diamonds). Number = 2.

• (ii) Event (Face card): Face cards are Jack, Queen, King. There are 3 face cards in each of the 4 suits. Total face cards = $3\times 4=12$. Number = 12.

• (iii) Event (Red face card): Red suits are Hearts and Diamonds. Each has 3 face cards (J, Q, K). Total red face cards = $3+3=6$. Number = 6.

• (iv) Event (Jack of hearts): There is only one Jack of Hearts. Number = 1.

• (v) Event (Spade): There are 13 spade cards. Number = 13.

• (vi) Event (Queen of diamonds): There is only one Queen of Diamonds. Number = 1.

Answer: (i) $\frac{1}{26}$, (ii) $\frac{3}{13}$, (iii) $\frac{3}{26}$, (iv) $\frac{1}{52}$, (v) $\frac{1}{4}$, (vi) $\frac{1}{52}$

15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

• Cards: {10♦, J♦, Q♦, K♦, A♦}. Total cards = 5.
• (i) Event (Queen is drawn): Favourable outcome = {Q♦}. Number = 1.

• (ii) After drawing the Queen and putting it aside:
• Remaining cards: {10♦, J♦, K♦, A♦}. Total remaining cards = 4.
• (a) Event (Ace is drawn second): Favourable outcome = {A♦}. Number = 1.

• (b) Event (Queen is drawn second): There are no Queens left in the deck. Number = 0.

Answer: (i) $\frac{1}{5}$, (ii) (a) $\frac{1}{4}$, (b) 0

16. 12 defective pens are accidentally mixed with 132 good ones. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

• Total pens: $12(\text{defective})+132(\text{good})=144$.
• Event (Good pen): Favourable outcomes = 132.

To simplify: Divide by 12: $\frac{11}{12}$. Answer: $\frac{11}{12}$

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?

• (i) Total bulbs: 20. Defective bulbs: 4.
• Event (Bulb is defective): Favourable outcomes = 4.

• (ii) After drawing one non-defective bulb and not replacing it:
• Total bulbs remaining: $20-1=19$.
• Non-defective bulbs remaining: $(20-4)-1=16-1=15$.
• Event (Second bulb is not defective): Favourable outcomes = 15.

Answer: (i) $\frac{1}{5}$, (ii) $\frac{15}{19}$

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

• Total discs: 90.
• (i) Event (Two-digit number): Numbers are from 10 to 90.
• Numbers from 1 to 9 are single-digit (9 numbers).
• Numbers from 10 to 90 are two-digit. Number of two-digit numbers = $90-9=81$.

• (ii) Event (Perfect square number): Perfect squares between 1 and 90 are: $1^2=1,2^2=4,3^2=9,4^2=16,5^2=25,6^2=36,7^2=49,8^2=64,9^2=81$. $({10}^2=100$ is too large).
• Favourable outcomes: {1, 4, 9, 16, 25, 36, 49, 64, 81}. Number = 9.

• (iii) Event (Number divisible by 5): Numbers are 5, 10, 15, ..., 90.
• This is an arithmetic progression. To find the count: $a_n=a_1+(n-1)d$.
• $90=5+(n-1)5$
• $85=(n-1)5$
• $17=n-1\Rightarrow n=18$.
• Alternatively, $90\mathrm{/}5=18$.
• Favourable outcomes = 18.

Answer: (i) $\frac{9}{10}$, (ii) $\frac{1}{10}$, (iii) $\frac{1}{5}$

19. A child has a die whose six faces show the letters as given below: A B C D E A The die is thrown once. What is the probability of getting (i) A? (ii) D?

• Sample Space (S): {A, B, C, D, E, A}. Total outcomes = 6.
• (i) Event (Getting A): The letter 'A' appears on 2 faces. Favourable outcomes = 2.

• (ii) Event (Getting D): The letter 'D' appears on 1 face. Favourable outcome = 1.

Answer: (i) $\frac{1}{3}$, (ii) $\frac{1}{6}$

20. Suppose you drop a die at random on a rectangular region. What is the probability that it will land inside the circle with diameter 1m?

• This is a geometric probability problem involving areas.
• Figure: A rectangle with a circle inside it. Let's assume the rectangle has dimensions $L\times W$. The problem doesn't give rectangle dimensions but says "rectangular region", implying the die lands within it. The circle has a diameter of 1m, so its radius is $r=0.5$ m.
• Area of circle: $A_{circle}=\pi r^2=\pi (0.5{)}^2=0.25\pi {\text{m}}^2$.
• Area of rectangle: This is needed for the total area. If we assume the diagram implies the rectangle's dimensions relative to the circle, or if the problem statement is incomplete without it, we'd need those dimensions. Assuming the question means the circle is inscribed or within a region of known area. If the rectangle's area is $A_{rect}$, then:

Without the rectangle's dimensions, we cannot solve this numerically. However, if we were given, say, a $2\text{m}\times 2\text{m}$ square region: $A_{rect}=2\times 2=4{\text{m}}^2$. $P(\text{Lands in circle})=\frac{0.25\pi }{4}=\frac{\pi }{16}$. (Note: This problem might be incomplete as stated without rectangle dimensions.)

21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it?

• Total pens: 144. Defective pens: 20. Good pens: $144-20=124$.
• (i) Event (She will buy it): Nuri buys only good pens. Favourable outcomes = 124.

Simplify by dividing by 4: $\frac{31}{36}$.

• (ii) Event (She will not buy it): Nuri does not buy defective pens. Favourable outcomes = 20.

Simplify by dividing by 4: $\frac{5}{36}$. (Alternatively: $1-P(\text{She will buy})=1-\frac{31}{36}=\frac{5}{36}$) Answer: (i) $\frac{31}{36}$, (ii) $\frac{5}{36}$

22. Refer to Example 13 (Tossing Two Dice). (i) Complete the following table:

• Sum 2: (1,1) - 1 outcome
• Sum 3: (1,2), (2,1) - 2 outcomes
• Sum 4: (1,3), (2,2), (3,1) - 3 outcomes
• Sum 5: (1,4), (2,3), (3,2), (4,1) - 4 outcomes
• Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes
• Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes
• Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes
• Sum 9: (3,6), (4,5), (5,4), (6,3) - 4 outcomes
• Sum 10: (4,6), (5,5), (6,4) - 3 outcomes
• Sum 11: (5,6), (6,5) - 2 outcomes
• Sum 12: (6,6) - 1 outcome Total outcomes = $1+2+3+4+5+6+5+4+3+2+1=36$.

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability $\frac{1}{11}$’. Do you agree with this argument? Justify your answer. Answer: No, I do not agree. Justification: The argument is incorrect because the outcomes (sums) are not equally likely. For example, a sum of 7 can be obtained in 6 ways ((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)), while a sum of 2 can only be obtained in 1 way ((1,1)). Since the number of favourable outcomes for each sum is different, their probabilities are different, not all equal to $\frac{1}{11}$.

23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

• Experiment: Tossing a coin 3 times.
• Sample Space (S): $2^3=8$ outcomes. {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
• Event (Hanif wins): {HHH, TTT}. Number of outcomes = 2.

• Event (Hanif loses): This is the complement of winning.
• Outcomes where Hanif loses: {HHT, HTH, THH, HTT, THT, TTH}. Number = 6.

• Alternatively, using complement rule:

Answer: $\frac{3}{4}$

24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?

• Experiment: Throwing a die twice.

• Sample Space (S): $6\times 6=36$ outcomes.

• (i) Event (5 will not come up either time): This means the outcome on the first throw is NOT 5, AND the outcome on the second throw is NOT 5.

• Number of outcomes for first throw (not 5) = 5 (i.e., 1, 2, 3, 4, 6).

• Number of outcomes for second throw (not 5) = 5 (i.e., 1, 2, 3, 4, 6).

• Number of favourable outcomes = $5\times 5=25$.

• (ii) Event (5 will come up at least once): This is the complement of "5 will not come up either time".
• Using complement rule:

(Direct method: Outcomes with at least one 5 are (1,5), (2,5), (3,5), (4,5), (6,5) [5 outcomes] and (5,1), (5,2), (5,3), (5,4), (5,6) [5 outcomes] and (5,5) [1 outcome]. Total = $5+5+1=11$). Answer: (i) $\frac{25}{36}$, (ii) $\frac{11}{36}$

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer. (i) Argument: If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $\frac{1}{3}$. Answer: Not correct. Reason: The possible outcomes are {HH, HT, TH, TT}. There are 4 equally likely outcomes. The outcome "one of each" includes both HT and TH. So, the outcomes are not just 3, and they are not equally likely in the way the argument implies (e.g., 'one of each' has probability 2/4).

(ii) Argument: If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is $\frac{1}{2}$. Answer: Correct. Reason: The sample space for a die is {1, 2, 3, 4, 5, 6}. The odd numbers are {1, 3, 5} (3 outcomes). The even numbers are {2, 4, 6} (3 outcomes). Since the number of odd outcomes equals the number of even outcomes, and all outcomes are equally likely, the probability of getting an odd number is $\frac{3}{6}=\frac{1}{2}$. This argument is sound.