GSEB Class 10 Maths Solutions Chapter 15 Probability Exercise 15.1

 

Question 1. Complete the following statements:
1. Probability of an event E + Probability of the event 'not E' = ........
2. The probability of an event that cannot happen is ........ Such an event is called ........
3. Probability of an event that is certain to happen is ........ Such an event is called ........
4. The sum of the probabilities of all the elementary events of an experiment is ........
5. The probability of an event is greater than or equal to ........ and less than or equal to ........
Answer:
1. Probability of an event E + Probability of the event 'not E' = 1
2. The probability of an event that cannot happen is 0. Such an event is called an impossible event.
3. Probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
4. The sum of the probabilities of all the elementary events of an experiment is 1.
5. The probability of an event is greater than or equal to 0 and less than or equal to 1.
In simple words: These statements explain the basic rules of probability. The chance of something happening plus the chance of it not happening always adds up to 1. If something can't happen, its chance is 0, and if it's guaranteed to happen, its chance is 1. The total chances of all possible simple events in an experiment is always 1, and any probability must fall between 0 and 1.

Exam Tip: Remember these fundamental definitions and values (0, 1) for impossible, sure, and complementary events. They are building blocks for all probability problems.

 

Question 2. For each of the following experiments, state whether the outcomes are equally likely or unequally likely. Give explanations where necessary.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shoot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Answer:
(i) Unequally likely outcome. The car might be old or new, in good condition or bad. Therefore, the chance of it starting or not starting is not necessarily equal.
(ii) Unequally likely outcome. The player's skill level affects the probability of making the shot or missing it. A skilled player has a higher chance of scoring.
(iii) Equally likely outcome. In a true-false question, there are only two options, and usually, each option has an equal chance of being correct or incorrect if guessing randomly.
(iv) Equally likely outcome. Historically and biologically, the probability of a baby being a boy or a girl is considered approximately equal, close to 50% for each.
In simple words: "Equally likely" means each result has the same chance of happening. "Unequally likely" means some results are more likely than others. Car starting and basketball shots depend on other things, so they're unequal. A true-false guess and a baby's gender are generally equal chances.

Exam Tip: To decide if outcomes are equally likely, consider if any external factors (skill, condition, bias) would make one outcome more probable than another. If not, they're likely equal.

 

Question 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?
Answer: In the experiment of tossing a coin, the possible outcomes (Head and Tail) are equally likely. This means each team gets an equal opportunity to win the toss and begin the game with the ball.
In simple words: Tossing a coin is fair because there's an equal chance of getting heads or tails. This gives both teams a 50-50 opportunity to start with the ball, so no team has an unfair advantage.

Exam Tip: When explaining fairness in probability, always emphasize that each outcome must have an equal chance of occurring. This removes bias and ensures impartiality.

 

Question 4. Which of the following cannot be the probability of an event?
(a) \( \frac {2}{3} \)
(b) -1.5
(c) 15%
(d) 0.7
Answer: (b) -1.5
In simple words: The probability of any event must always be a number between 0 and 1. It cannot be negative, and it cannot be greater than 1. So, -1.5 cannot be a probability.

Exam Tip: Always remember that probability \( P(E) \) must satisfy \( 0 \le P(E) \le 1 \). Any value outside this range cannot be a valid probability.

 

Question 5. If P(E) = 0.05, what is the probability of 'not E'?
Answer: We know that the sum of the probability of an event and the probability of its complement (not happening) is always 1.
\( P(E) + P(\text{not E}) = 1 \)
Given \( P(E) = 0.05 \).
\( 0.05 + P(\text{not E}) = 1 \)
\( P(\text{not E}) = 1 - 0.05 \)
\( P(\text{not E}) = 0.95 \)
In simple words: If the chance of something happening is 0.05, then the chance of it not happening is found by subtracting 0.05 from 1, which gives 0.95.

Exam Tip: The formula \( P(E) + P(\text{not E}) = 1 \) (or \( P(E') = 1 - P(E) \)) is essential. Use it whenever you need to find the probability of an event not occurring.

 

Question 6. A bag contains lemon-flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
1. an orange-flavored candy?
2. a lemon-flavored candy? (CBSE 2012)
Answer:
1. As the bag contains only lemon-flavored candies, the probability that she takes out an orange-flavored candy is '0'. This is considered an impossible event.
2. As the bag contains only lemon-flavored candies, the probability that she takes out a lemon-flavored candy is '1'. This is considered a sure or certain event.
In simple words: If a bag only has lemon candies, the chance of picking an orange candy is zero because there aren't any. The chance of picking a lemon candy is 1, meaning it's guaranteed because all the candies are lemon.

Exam Tip: Always analyze the contents of the sample space (the bag in this case). If an event's outcome is not in the sample space, its probability is 0. If all outcomes in the sample space match the event, its probability is 1.

 

Question 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?
Answer: Let E be the event that 2 students do not have the same birthday.
Let E' be the event that 2 students have the same birthday.
We are given \( P(E) = 0.992 \).
We know that \( P(E) + P(E') = 1 \).
So, \( P(\text{2 students have the same birthday}) = 1 - P(\text{2 students not having the same birthday}) \)
\( P(\text{2 students have the same birthday}) = 1 - 0.992 \)
\( P(\text{2 students have the same birthday}) = 0.008 \)
In simple words: If there's a 0.992 chance that two students don't share a birthday, then the chance that they do share a birthday is 1 minus 0.992, which is 0.008.

Exam Tip: This problem is a direct application of the complement rule. Recognize when to use \( P(E') = 1 - P(E) \) to simplify calculations for "not" events.

 

Question 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?
Answer:
Number of red balls = 3
Number of black balls = 5
Total number of balls = \( 3 + 5 = 8 \)
(i) Probability of drawing a red ball \( P(\text{red ball}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{8} \)
(ii) Probability of not drawing a red ball \( P(\text{not red ball}) = 1 - P(\text{red ball}) = 1 - \frac{3}{8} = \frac{8-3}{8} = \frac{5}{8} \)
In simple words: There are 8 balls in total. If 3 are red, the chance of picking a red ball is 3 out of 8. If 5 balls are not red, the chance of picking a non-red ball is 5 out of 8, or you can subtract the red ball probability from 1.

Exam Tip: Clearly state the total number of outcomes and the number of favorable outcomes for each event. Use the complement rule for "not" events to save time and ensure accuracy.

 

Question 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 is taken out will be (i) red (ii) white (iii) not green?
Answer:
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles = \( 5 + 8 + 4 = 17 \)
Therefore, the number of all possible outcomes = 17
(i) Probability of drawing a red marble \( P(\text{red}) = \frac{\text{Number of outcomes favourable to red}}{\text{Number of all possible outcomes}} = \frac{5}{17} \)
(ii) Probability of drawing a white marble \( P(\text{white}) = \frac{\text{Number of outcomes favourable to white}}{\text{Number of all possible outcomes}} = \frac{8}{17} \)
(iii) Probability of not drawing a green marble \( P(\text{not green}) = 1 - P(\text{green}) \)
First, find \( P(\text{green}) \):
\( P(\text{green}) = \frac{\text{Number of outcomes favourable to green}}{\text{Number of all possible outcomes}} = \frac{4}{17} \)
So, \( P(\text{not green}) = 1 - \frac{4}{17} = \frac{17-4}{17} = \frac{13}{17} \)
In simple words: Add up all the marbles to find the total. Then, to get the chance of picking a certain color, divide the number of marbles of that color by the total number. To find the chance of "not green," subtract the chance of "green" from 1.

Exam Tip: When dealing with multiple categories, always start by finding the total number of items. For "not X" probabilities, using the complement rule \( 1 - P(X) \) is often more efficient than counting all non-X items.

 

Question 10. A piggy bank contains a hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 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 of a 50 p coin? (ii) will not be a Rs 5 coin?
Answer:
Number of 50 p coins in the piggy bank = 100
Number of Rs 1 coins in the piggy bank = 50
Number of Rs 2 coins = 20
Number of Rs 5 coins = 10
Total number of coins = \( 100 + 50 + 20 + 10 = 180 \)
(i) Probability of the coin being a 50 p coin \( P(\text{50 p coin}) = \frac{100}{180} = \frac{5}{9} \)
(ii) Probability of the coin not being a Rs 5 coin \( P(\text{not Rs 5 coin}) = 1 - P(\text{Rs 5 coin}) \)
First, find \( P(\text{Rs 5 coin}) \):
\( P(\text{Rs 5 coin}) = \frac{10}{180} = \frac{1}{18} \)
So, \( P(\text{not Rs 5 coin}) = 1 - \frac{1}{18} = \frac{18-1}{18} = \frac{17}{18} \)
In simple words: First, count all the coins to get the total. For part (i), divide the number of 50 p coins by the total. For part (ii), find the chance of picking a Rs 5 coin, then subtract that from 1 to get the chance of not picking a Rs 5 coin.

Exam Tip: Carefully identify the number of each type of item and the total. Ensure all currency symbols are correctly converted and calculations are simplified to their lowest terms.

 

Question 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 (see figure). What is the probability that the fish taken out is a male fish?
Answer:
Number of male fish = 5
Number of female fish = 8
Total number of fishes = \( 5 + 8 = 13 \)
Probability that the fish taken out is a male fish \( P(\text{male fish}) = \frac{\text{Number of male fish}}{\text{Total number of fishes}} = \frac{5}{13} \)
In simple words: To find the chance of getting a male fish, you divide the count of male fish by the total count of all fish in the tank.

Exam Tip: When given a problem with different categories of items, always sum them up to find the total sample space before calculating individual probabilities.

 

Question 12. A game of chance consists of spinning an arrow that comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure), and these are equally likely outcomes. What is the probability that it will point at
1. 8?
2. an odd number?
3. a number greater than 2?
4. a number less than 9?
Answer:
Total number of outcomes = 8 (since the numbers are 1, 2, 3, 4, 5, 6, 7, 8)
1. Probability of pointing at 8 \( P(8) = \frac{\text{Number of times 8 appears}}{\text{Total number of outcomes}} = \frac{1}{8} \)
2. Odd numbers are 1, 3, 5, 7. There are 4 odd numbers.
Probability of pointing at an odd number \( P(\text{an odd number}) = \frac{\text{Number of odd numbers}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2} \)
3. Numbers greater than 2 are 3, 4, 5, 6, 7, 8. There are 6 such numbers.
Probability of pointing at a number greater than 2 \( P(\text{number greater than 2}) = \frac{\text{Number of numbers greater than 2}}{\text{Total number of outcomes}} = \frac{6}{8} = \frac{3}{4} \)
4. All the numbers (1, 2, 3, 4, 5, 6, 7, 8) are less than 9. There are 8 such numbers.
Probability of pointing at a number less than 9 \( P(\text{number less than 9}) = \frac{\text{Number of numbers less than 9}}{\text{Total number of outcomes}} = \frac{8}{8} = 1 \) (This is a certain event).
In simple words: The spinner has 8 equal parts. To find any probability, count how many numbers match your condition and divide by 8. For example, there's only one '8', so the chance is 1/8. There are four odd numbers, so the chance is 4/8 or 1/2. All numbers are less than 9, so that's a sure thing, probability 1.

Exam Tip: When outcomes are equally likely, probability is simply the ratio of favorable outcomes to the total number of outcomes. Clearly list the favorable outcomes for each sub-part before calculating.

 

Question 13. A die is thrown once. Find the probability of getting
1. a prime number
2. a number lying between 2 and 6
3. an odd number
Answer: When a die is thrown once, the number of all possible outcomes are {1, 2, 3, 4, 5, 6}. The total number of outcomes is 6.
1. Prime numbers among {1, 2, 3, 4, 5, 6} are 2, 3, 5. There are 3 prime numbers.
Probability of getting a prime number \( P(\text{a prime number}) = \frac{\text{Number of prime numbers}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2} \)
2. Numbers lying between 2 and 6 are 3, 4, 5. There are 3 such numbers.
Probability of getting a number lying between 2 and 6 \( P(\text{a number lying between 2 and 6}) = \frac{\text{Number of numbers between 2 and 6}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2} \)
3. Odd numbers among {1, 2, 3, 4, 5, 6} are 1, 3, 5. There are 3 odd numbers.
Probability of getting an odd number \( P(\text{an odd number}) = \frac{\text{Number of odd numbers}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2} \)
In simple words: When you roll a die, there are 6 possible results. For each part of the question, count how many of those 6 results fit the rule, then divide that count by 6 to get the probability. For prime, odd, or numbers between 2 and 6, there are 3 choices each, making the chance 3/6 or 1/2.

Exam Tip: Remember the sample space for a single die roll. Carefully list the favorable outcomes for each event (prime, between 2 and 6, odd) to avoid errors in counting.

 

Question 14. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
1. a king of red colour
2. a face card
3. a red face card
4. the jack of hearts
5. a spade
6. the queen of diamonds. (CSBE 2012)
Answer: Total number of cards in a deck = 52
1. There are 2 kings of red colour (King of Hearts and King of Diamonds).
Probability of getting a king of red colour \( P(\text{a king of red colour}) = \frac{2}{52} = \frac{1}{26} \)
2. There are 12 face cards in a deck (3 face cards per suit x 4 suits = 12). (Jack, Queen, King of each suit).
Probability of getting a face card \( P(\text{a face card}) = \frac{12}{52} = \frac{3}{13} \)
3. There are 6 red face cards (King, Queen, Jack of Hearts and King, Queen, Jack of Diamonds).
Probability of getting a red face card \( P(\text{a red face card}) = \frac{6}{52} = \frac{3}{26} \)
4. There is only 1 jack of hearts in a deck.
Probability of getting the jack of hearts \( P(\text{the jack of hearts}) = \frac{1}{52} \)
5. There are 13 spades in a deck.
Probability of getting a spade \( P(\text{a spade}) = \frac{13}{52} = \frac{1}{4} \)
6. There is only 1 queen of diamonds in a deck.
Probability of getting the queen of diamonds \( P(\text{the queen of diamonds}) = \frac{1}{52} \)
In simple words: A deck has 52 cards. For each part, count how many cards fit the description and divide that number by 52. Remember kings, queens, and jacks are face cards. Red kings are two, face cards are twelve, red face cards are six. There's only one Jack of Hearts and one Queen of Diamonds. Spades are a full suit of thirteen cards.

Exam Tip: Familiarize yourself with the structure of a standard 52-card deck (suits, number of cards per suit, face cards, specific cards). This knowledge is critical for solving card-related probability problems.

 

Question 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?
Answer:
Initially, the cards are: Ten, Jack, Queen, King, and Ace of Diamonds.
Total number of cards = 5
(i) Probability that the card is the queen: There is only one queen in this set of 5 cards.
\( P(\text{queen}) = \frac{1}{5} \)
(ii) If the queen is drawn and put aside, the remaining cards are: Ten, Jack, King, and Ace of Diamonds.
Now, the total number of cards left = 4
(a) Probability that the second card picked up is an ace: There is only one ace among the remaining 4 cards.
\( P(\text{an ace}) = \frac{1}{4} \)
(b) Probability that the second card picked up is a queen: Since the queen was already removed and not replaced, there are no queens left among the remaining 4 cards.
\( P(\text{a queen}) = \frac{0}{4} = 0 \)
In simple words: At first, with 5 cards, the chance of picking the queen is 1 out of 5. If you take the queen out, only 4 cards are left. Now, the chance of picking an ace is 1 out of 4. The chance of picking another queen is 0 because it's already gone.

Exam Tip: For problems involving drawing without replacement, remember that the total number of outcomes (and sometimes favorable outcomes) changes after each draw. Adjust the denominator and numerator accordingly for subsequent probabilities.

 

Question 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen is taken out is a good one. (CBSE 2012)
Answer:
Number of defective pens = 12
Number of good pens = 132
Total number of pens = \( 12 + 132 = 144 \)
Probability that the pen taken out is a good one \( P(\text{Good one}) = \frac{\text{Number of good pens}}{\text{Total number of pens}} = \frac{132}{144} \)
To simplify the fraction, divide both numerator and denominator by their greatest common divisor. Both are divisible by 12.
\( \frac{132 \div 12}{144 \div 12} = \frac{11}{12} \)
In simple words: First, add the defective and good pens to find the total count. Then, to get the chance of picking a good pen, divide the number of good pens by the total number of pens, and simplify the fraction.

Exam Tip: Always calculate the total number of items first. Then, make sure to simplify the probability fraction to its lowest terms, as this is often required in final answers.

 

Question 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 not defective?
Answer:
Total number of bulbs = 20
Number of defective bulbs = 4
Number of non-defective (good) bulbs = \( 20 - 4 = 16 \)
(i) Probability of drawing a defective bulb \( P(\text{defective bulb}) = \frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}} = \frac{4}{20} = \frac{1}{5} \)
(ii) Now, suppose the bulb drawn in (i) was NOT defective and was NOT replaced.
This means one good bulb has been removed.
Total number of bulbs left = \( 20 - 1 = 19 \)
Number of defective bulbs still = 4
Number of non-defective (good) bulbs left = \( 16 - 1 = 15 \)
Probability that this (second) bulb is not defective \( P(\text{not defective}) = \frac{\text{Number of non-defective bulbs left}}{\text{Total number of bulbs left}} = \frac{15}{19} \)
In simple words: (i) With 4 bad bulbs out of 20, the chance of picking a bad one is 4/20, or 1/5. (ii) If a good bulb is taken out and not put back, there are now 19 bulbs left, with 15 of them being good. So, the chance of picking another good bulb is 15/19.

Exam Tip: Pay close attention to whether items are replaced or not. "Not replaced" means both the total number of outcomes and sometimes the number of favorable outcomes change for subsequent draws.

 

Question 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:
1. a two-digit number.
2. a perfect square number.
3. a number divisible by 5.
Answer:
Total number of discs = 90 (numbered from 1 to 90)
1. Two-digit numbers are from 10 to 90.
Count of two-digit numbers = \( 90 - 10 + 1 = 81 \). (Alternatively, total numbers 90, single-digit numbers 1 to 9 (9 numbers). So, \( 90 - 9 = 81 \)).
Favorable cases for a two-digit number are 10, 11, 12, 13, ..., 90, which is 81 numbers.
Probability of drawing a two-digit number \( P(\text{a two-digit number}) = \frac{81}{90} = \frac{9}{10} \)
2. Perfect square numbers between 1 and 90 are \( 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81 \). There are 9 perfect square numbers.
Probability of drawing a perfect square number \( P(\text{a perfect square number}) = \frac{9}{90} = \frac{1}{10} \)
3. Numbers divisible by 5 between 1 and 90 are 5, 10, 15, 20, ..., 90.
To count them, divide the last number by 5: \( \frac{90}{5} = 18 \). So there are 18 such numbers.
Probability of drawing a number divisible by 5 \( P(\text{a number divisible by 5}) = \frac{18}{90} = \frac{1}{5} \)
In simple words: You have 90 numbered discs. (1) To find the chance of picking a two-digit number, count all numbers from 10 to 90 (there are 81) and divide by 90. (2) For perfect squares, list all numbers that are a number multiplied by itself (like 1, 4, 9, etc.) up to 90 (there are 9) and divide by 90. (3) For numbers divisible by 5, count every fifth number (5, 10, 15, ...) up to 90 (there are 18) and divide by 90.

Exam Tip: For number-related probability questions, carefully list or count the favorable outcomes. For sequences like multiples, using division can quickly determine the count. Always simplify fractions.

 

Question 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?
Answer:
The six faces of the die show the letters: A, B, C, D, E, A.
Total number of faces = 6
(i) Probability of getting A: The letter 'A' appears on 2 faces.
\( P(\text{A}) = \frac{\text{Number of faces with A}}{\text{Total number of faces}} = \frac{2}{6} = \frac{1}{3} \)
(ii) Probability of getting D: The letter 'D' appears on 1 face.
\( P(\text{D}) = \frac{\text{Number of faces with D}}{\text{Total number of faces}} = \frac{1}{6} \)
In simple words: This special die has 6 sides. To find the chance of rolling 'A', count how many 'A's there are (2) and divide by 6. To find the chance of rolling 'D', count how many 'D's there are (1) and divide by 6.

Exam Tip: When dealing with custom dice or objects, always identify the complete set of possible outcomes (sample space) and then count the occurrences of the specific event. The total number of outcomes is the denominator.

 

Question 20. Suppose you drop a die at random on the rectangular region shown in Figure. What is the probability that it will land inside the circle with diameter 1 m?
Answer:
The rectangular region has a length of 3 m and a width of 2 m.
Area of the rectangular region = Length \( \times \) Width = \( 3 \text{ m} \times 2 \text{ m} = 6 \text{ m}^2 \).
The circle has a diameter of 1 m, so its radius \( r = \frac{1}{2} \text{ m} = 0.5 \text{ m} \).
Area of the circle = \( \pi r^2 = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4} \text{ m}^2 \).
The probability that the die will land inside the circle is the ratio of the area of the circle to the area of the rectangular region.
\( P(\text{die will land inside the circle}) = \frac{\text{Area of circle}}{\text{Area of rectangular region}} = \frac{\frac{\pi}{4}}{6} = \frac{\pi}{4 \times 6} = \frac{\pi}{24} \)
In simple words: First, calculate the area of the whole rectangular shape (length times width). Then, figure out the area of the circle using its radius. The chance of the die landing inside the circle is the circle's area divided by the rectangle's total area.

Exam Tip: For geometric probability problems, the probability is usually calculated as the ratio of the favorable area (or volume) to the total area (or volume) of the sample space. Ensure correct area formulas and unit consistency.

 

Question 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?
Answer:
Total number of ball pens = 144
Number of defective ball pens = 20
Number of good ball pens = Total pens - Defective pens = \( 144 - 20 = 124 \)
(i) Probability that Nuri will buy the pen: Nuri buys a pen only if it is good.
\( P(\text{she will buy it}) = P(\text{good pen}) = \frac{\text{Number of good ball pens}}{\text{Total number of ball pens}} = \frac{124}{144} \)
To simplify the fraction, divide both numerator and denominator by their greatest common divisor. Both are divisible by 4.
\( \frac{124 \div 4}{144 \div 4} = \frac{31}{36} \)
(ii) Probability that Nuri will not buy the pen: Nuri will not buy if the pen is defective.
\( P(\text{she will not buy it}) = P(\text{defective pen}) = \frac{\text{Number of defective ball pens}}{\text{Total number of ball pens}} = \frac{20}{144} \)
To simplify the fraction, divide both numerator and denominator by their greatest common divisor. Both are divisible by 4.
\( \frac{20 \div 4}{144 \div 4} = \frac{5}{36} \)
Alternatively, using the complement rule for (ii):
\( P(\text{she will not buy it}) = 1 - P(\text{she will buy it}) = 1 - \frac{31}{36} = \frac{36-31}{36} = \frac{5}{36} \)
In simple words: First, find out how many pens are good and how many are bad from the total of 144 pens. (i) Nuri only buys good pens, so the chance she buys one is the number of good pens divided by the total. (ii) Nuri won't buy a bad pen, so the chance she won't buy is the number of bad pens divided by the total. You can also get this by subtracting the 'buy' probability from 1.

Exam Tip: Understand the conditions for buying/not buying to correctly identify the favorable outcomes. The complement rule \( P(\text{not A}) = 1 - P(A) \) is often useful and can provide a quick check for your calculations.

 

Question 22.
1. Two dice, one blue and one grey are thrown at the same time. Complete the following table:
2. 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:
1. When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \).
The sample space S consists of 36 ordered pairs:
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)}
We need to find the number of favorable cases for each sum:
P(a total of 2) = \( \frac{1}{36} \) (Favorable case: (1, 1))
P(a total of 3) = \( \frac{2}{36} \) (Favorable cases: (1, 2), (2, 1))
P(a total of 4) = \( \frac{3}{36} \) (Favorable cases: (1, 3), (3, 1), (2, 2))
P(a total of 5) = \( \frac{4}{36} \) (Favorable cases: (1, 4), (4, 1), (2, 3), (3, 2))
P(a total of 6) = \( \frac{5}{36} \) (Favorable cases: (1, 5), (5, 1), (2, 4), (4, 2), (3, 3))
P(a total of 7) = \( \frac{6}{36} \) (Favorable cases: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3))
P(a total of 8) = \( \frac{5}{36} \) (Favorable cases: (2, 6), (6, 2), (3, 5), (5, 3), (4, 4))
P(a total of 9) = \( \frac{4}{36} \) (Favorable cases: (3, 6), (6, 3), (4, 5), (5, 4))
P(a total of 10) = \( \frac{3}{36} \) (Favorable cases: (4, 6), (6, 4), (5, 5))
P(a total of 11) = \( \frac{2}{36} \) (Favorable cases: (5, 6), (6, 5))
P(a total of 12) = \( \frac{1}{36} \) (Favorable case: (6, 6))
Now, completing the table:

2. No, the student's argument is incorrect. While there are 11 possible sums when rolling two dice (from 2 to 12), these outcomes are not equally likely. As shown in the table above, the probability for each sum is different, ranging from \( \frac{1}{36} \) for sums of 2 or 12 to \( \frac{6}{36} \) for a sum of 7. For outcomes to have equal probability, each outcome must have the same number of favorable combinations, which is not the case here.
In simple words: (1) When you roll two dice, there are 36 possible pairs of numbers. You need to count how many pairs add up to each total (like 2, 3, 4, etc.) and then divide by 36 to get the probability. (2) The student is wrong because sums like 2 (only 1+1) and 7 (many ways like 1+6, 2+5) don't have the same number of combinations, so they don't have the same chance of happening.

Exam Tip: When dealing with two dice, always remember that there are 36 possible outcomes. Listing the sample space or understanding the number of combinations for each sum is crucial for calculating probabilities accurately. Outcomes are only equally likely if each one has the same chance of occurring.

 

Question 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.
Answer: When a coin is tossed 3 times, the total possible outcomes are:
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
The total number of outcomes \( n(S) = 8 \).
Hanif wins if all the tosses give the same result, meaning three heads (HHH) or three tails (TTT).
Favorable outcomes for Hanif winning = {HHH, TTT}
Number of favorable outcomes for Hanif winning = 2
Probability that Hanif wins \( P(\text{Hanif wins}) = \frac{\text{Number of winning outcomes}}{\text{Total number of outcomes}} = \frac{2}{8} = \frac{1}{4} \)
Hanif loses if he does not win. The probability that Hanif loses is the complement of Hanif winning.
Probability that Hanif loses \( P(\text{Hanif loses}) = 1 - P(\text{Hanif wins}) \)
\( P(\text{Hanif loses}) = 1 - \frac{1}{4} = \frac{4-1}{4} = \frac{3}{4} \)
In simple words: If you flip a coin three times, there are 8 possible results. Hanif wins if he gets all heads or all tails (2 ways). So, the chance of him winning is 2 out of 8, or 1/4. The chance of him losing is 1 minus his winning chance, which is 3/4.

Exam Tip: Clearly list all possible outcomes in the sample space for sequential events. Identify the favorable outcomes for the specified event (Hanif wins) and then use the complement rule for "loses" or "not" events.

 

Question 24. A die is thrown twice. What is the probability that
1. 5 will not come up either time ?
2. 5 will come up at least once ?
[Hint. Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
Answer: A die is thrown twice, so the total number of ways (outcomes) is \( 6 \times 6 = 36 \).
First, let's list the outcomes where '5' comes up at least once.
These are the outcomes where 5 appears on the first throw or the second throw or both:
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5) --- 6 outcomes where 5 is on the second die.
(5, 1), (5, 2), (5, 3), (5, 4), (5, 6) --- 5 additional outcomes where 5 is on the first die (excluding (5,5) which is already counted).
So, favorable outcomes where 5 comes up at least once = \( 6 + 5 = 11 \) ways.
1. Probability that 5 will not come up either time:
This is the complement of "5 will come up at least once".
Number of unfavorable ways (where 5 does not come up either time) = Total ways - (5 comes up at least once) = \( 36 - 11 = 25 \).
\( P(\text{5 will not come up either time}) = \frac{25}{36} \)
2. Probability that 5 will come up at least once:
Number of favorable outcomes for this event = 11 ways (as counted above).
\( P(\text{5 will come up at least once}) = \frac{11}{36} \)
In simple words: When a die is rolled twice, there are 36 total possibilities. First, figure out all the times a '5' shows up (like 5,1; 1,5; 5,5). There are 11 such outcomes. (1) To find the chance that '5' doesn't show up at all, subtract those 11 outcomes from 36, leaving 25, so the chance is 25/36. (2) The chance that '5' shows up at least once is simply those 11 outcomes divided by 36.

Exam Tip: For problems involving "at least once," it's often easier to calculate the probability of the complementary event ("not at all") and subtract it from 1. Listing specific favorable outcomes helps avoid counting errors for either method.

 

Question 25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
1. 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 \( \frac {1}{3} \)
2. 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:
1. The argument is incorrect.
When two coins are tossed simultaneously, the actual sample space is S = {HH, HT, TH, TT}. The total number of outcomes \( n(S) = 4 \).
- Probability of two heads \( P(\text{2 Heads}) = \frac{1}{4} \) (outcome: HH)
- Probability of two tails \( P(\text{2 Tails}) = \frac{1}{4} \) (outcome: TT)
- Probability of one of each \( P(\text{one each}) = \frac{2}{4} = \frac{1}{2} \) (outcomes: HT, TH)
Since the probabilities for "two heads," "two tails," and "one of each" are not all \( \frac{1}{3} \), the outcomes are not equally likely. Specifically, "one of each" has twice the probability of "two heads" or "two tails" because there are two distinct ways to get one of each (HT and TH).
2. The argument is correct.
When a die is thrown, the sample space is S = {1, 2, 3, 4, 5, 6}. The total number of outcomes \( n(S) = 6 \).
- Odd numbers are {1, 3, 5}. There are 3 odd numbers.
Probability of getting an odd number \( P(\text{odd numbers}) = \frac{3}{6} = \frac{1}{2} \)
- Even numbers are {2, 4, 6}. There are 3 even numbers.
Probability of getting an even number \( P(\text{even numbers}) = \frac{3}{6} = \frac{1}{2} \)
Since the number of odd outcomes and even outcomes are equal, each has a probability of \( \frac{1}{2} \). The statement correctly recognizes that these two categories are equally likely.
In simple words: (1) The first argument is wrong. When tossing two coins, getting "one of each" is more likely than getting "two heads" or "two tails" because there are two ways to get one of each (HT and TH). So, the chances are not all 1/3. (2) The second argument is correct. When rolling a die, there are 3 odd numbers and 3 even numbers. This means the chance of getting an odd number is 3 out of 6, which is 1/2.

Exam Tip: For arguments about probability, always define the full sample space. If outcomes are grouped (like "one of each"), carefully count the individual events that contribute to that group. Outcomes are only equally likely if each individual combination has the same chance.