Worksheet
Probability is a measure of the likelihood that an event will occur, calculated as the ratio of favorable outcomes to the total number of possible outcomes.
Probability - Practice Worksheet
Strengthen your foundation with key concepts and basic applications.
This worksheet covers essential long-answer questions to help you build confidence in Probability from Mathematics for Class X (Mathematics).
Basic comprehension exercises
Strengthen your understanding with fundamental questions about the chapter.
Questions
Define probability and explain its theoretical approach with an example.
Think about the basic definition of probability and how it applies to simple experiments like tossing a coin or rolling a die.
Solution
Probability is a measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. The theoretical approach to probability assumes that all outcomes of an experiment are equally likely. For example, when a fair coin is tossed, there are two possible outcomes: heads or tails. Since the coin is fair, the probability of getting heads is 1/2, and the same is true for tails. This approach is based on the assumption of equally likely outcomes and does not require actual experimentation. The formula for theoretical probability is P(E) = Number of favorable outcomes / Total number of possible outcomes. This method is widely used in situations where the outcomes are known and can be clearly defined, such as in games of chance or in predicting the outcome of a die roll.
What is the difference between empirical and theoretical probability? Provide examples for each.
Consider how probability is calculated in real-life experiments versus ideal scenarios.
Solution
Empirical probability is based on actual experiments and observations, whereas theoretical probability is based on the assumption of equally likely outcomes. Empirical probability is calculated as the ratio of the number of times an event occurs to the total number of trials. For example, if you flip a coin 100 times and get heads 55 times, the empirical probability of getting heads is 55/100 = 0.55. Theoretical probability, on the other hand, is calculated based on the possible outcomes. For a fair coin, the theoretical probability of getting heads is 1/2 = 0.5. Empirical probability can vary with the number of trials and may not match the theoretical probability exactly, especially with a small number of trials. However, as the number of trials increases, the empirical probability tends to converge to the theoretical probability. This is known as the Law of Large Numbers.
Explain the concept of complementary events in probability with an example.
Think about events that cover all possible outcomes and are mutually exclusive.
Solution
Complementary events in probability are two events that are mutually exclusive and exhaustive, meaning that one event must occur and the other cannot. The sum of their probabilities is always 1. For example, when rolling a die, the event of getting an even number and the event of getting an odd number are complementary. If E is the event of getting an even number (2, 4, 6), then the complementary event E' is getting an odd number (1, 3, 5). The probability of E is 3/6 = 0.5, and the probability of E' is also 3/6 = 0.5. Together, their probabilities add up to 1. Complementary events are useful in probability calculations because knowing the probability of one event allows you to easily find the probability of the other. For instance, if the probability of raining today is 0.3, then the probability of it not raining is 1 - 0.3 = 0.7.
A bag contains 5 red, 3 blue, and 2 green balls. If one ball is drawn at random, what is the probability that it is not red? Explain your steps.
Consider using the complementary probability approach for an alternative solution.
Solution
To find the probability that a drawn ball is not red, we first determine the total number of balls and the number of balls that are not red. The bag contains 5 red, 3 blue, and 2 green balls, so the total number of balls is 5 + 3 + 2 = 10. The number of balls that are not red is the sum of blue and green balls, which is 3 + 2 = 5. The probability of drawing a ball that is not red is the ratio of the number of non-red balls to the total number of balls, which is 5/10 = 0.5. This can also be calculated using the concept of complementary events: the probability of not drawing a red ball is 1 minus the probability of drawing a red ball. The probability of drawing a red ball is 5/10 = 0.5, so the probability of not drawing a red ball is 1 - 0.5 = 0.5. Both methods yield the same result, confirming the answer.
Two dice are thrown simultaneously. What is the probability that the sum of the numbers on the two dice is 7? Explain your reasoning.
List all possible pairs of dice rolls that add up to 7 to find the number of favorable outcomes.
Solution
When two dice are thrown simultaneously, there are 6 x 6 = 36 possible outcomes, as each die has 6 faces. The favorable outcomes for the sum to be 7 are the pairs (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 such pairs. Therefore, the probability of the sum being 7 is the number of favorable outcomes divided by the total number of possible outcomes, which is 6/36 = 1/6 ≈ 0.1667. This calculation assumes that the dice are fair and that all outcomes are equally likely. The probability can also be visualized by listing all possible outcomes in a grid and counting the favorable ones. This problem illustrates how probability can be determined by enumerating all possible outcomes and identifying those that meet the desired condition.
What is the probability of drawing a king or a queen from a well-shuffled deck of 52 cards? Explain your steps.
Count the total number of kings and queens in the deck to find the number of favorable outcomes.
Solution
A standard deck of 52 cards contains 4 kings and 4 queens, making a total of 8 cards that are either kings or queens. The probability of drawing a king or a queen is the ratio of the number of favorable cards to the total number of cards in the deck. Therefore, the probability is 8/52, which simplifies to 2/13 ≈ 0.1538. This calculation assumes that each card has an equal chance of being drawn, which is true for a well-shuffled deck. The problem demonstrates the application of probability in card games and the importance of understanding the composition of a deck. It also shows how to combine the probabilities of two distinct events (drawing a king and drawing a queen) when they are mutually exclusive, meaning that a single card cannot be both a king and a queen at the same time.
A box contains 10 bulbs, of which 4 are defective. If one bulb is drawn at random, what is the probability that it is not defective? Explain your reasoning.
Subtract the number of defective bulbs from the total to find the number of non-defective bulbs.
Solution
The box contains 10 bulbs in total, with 4 being defective and 6 being non-defective. The probability of drawing a non-defective bulb is the ratio of the number of non-defective bulbs to the total number of bulbs, which is 6/10 = 0.6. This means there is a 60% chance that the drawn bulb will not be defective. The problem illustrates the basic principle of probability, where the likelihood of an event is determined by the proportion of favorable outcomes to the total possible outcomes. It also highlights the practical application of probability in quality control and inspection processes, where understanding the probability of defects can help in making informed decisions.
Explain the concept of equally likely outcomes in probability with an example.
Consider simple experiments like tossing a coin or rolling a die where each outcome has the same chance.
Solution
Equally likely outcomes in probability refer to outcomes that have the same chance of occurring. This concept is fundamental to the theoretical approach of probability, where it is assumed that all possible outcomes of an experiment are equally likely. For example, when rolling a fair six-sided die, each face (1, 2, 3, 4, 5, 6) has an equal probability of 1/6. This assumption allows for the calculation of probabilities by simply counting the number of favorable outcomes and dividing by the total number of possible outcomes. However, in real-world scenarios, outcomes may not always be equally likely, such as when a die is biased. The concept of equally likely outcomes is crucial for simplifying probability problems and is widely used in games of chance, statistical experiments, and predictive models.
A card is drawn from a well-shuffled deck of 52 cards. What is the probability that it is a face card or a spade? Explain your steps.
Be careful not to double-count the face cards that are also spades.
Solution
A standard deck of 52 cards contains 12 face cards (4 kings, 4 queens, 4 jacks) and 13 spades. However, 3 of the face cards are also spades (the king, queen, and jack of spades), so these are counted in both groups. To avoid double-counting, we use the principle of inclusion-exclusion. The total number of favorable outcomes is the number of face cards plus the number of spades minus the number of cards that are both face cards and spades: 12 + 13 - 3 = 22. Therefore, the probability of drawing a face card or a spade is 22/52, which simplifies to 11/26 ≈ 0.4231. This problem demonstrates how to handle overlapping events in probability calculations and the importance of adjusting for double-counting when combining probabilities.
A jar contains 24 marbles, some green and some blue. If the probability of drawing a green marble is 2/3, how many green marbles are in the jar? Explain your reasoning.
Use the given probability to set up a proportion and solve for the unknown quantity.
Solution
Given that the probability of drawing a green marble is 2/3, we can set up the equation: (Number of green marbles) / (Total number of marbles) = 2/3. The total number of marbles is 24, so the equation becomes: Number of green marbles / 24 = 2/3. To find the number of green marbles, we multiply both sides of the equation by 24: Number of green marbles = (2/3) * 24 = 16. Therefore, there are 16 green marbles in the jar. This problem illustrates how probability can be used to determine the composition of a set when the total number of items and the probability of a particular event are known. It also shows the application of basic algebraic techniques in solving probability problems.
Probability - Mastery Worksheet
Advance your understanding through integrative and tricky questions.
This worksheet challenges you with deeper, multi-concept long-answer questions from Probability to prepare for higher-weightage questions in Class X Mathematics.
Intermediate analysis exercises
Deepen your understanding with analytical questions about themes and characters.
Questions
A bag contains 5 red, 3 blue, and 2 green marbles. If two marbles are drawn at random without replacement, what is the probability that both are red?
Consider the change in total number of marbles and red marbles after the first draw.
Solution
First, calculate the total number of marbles: 5 red + 3 blue + 2 green = 10 marbles. The probability of drawing the first red marble is 5/10. After drawing one red marble, there are now 4 red marbles left and 9 marbles in total. So, the probability of drawing a second red marble is 4/9. The combined probability is (5/10) * (4/9) = 20/90 = 2/9.
Compare and contrast empirical probability and theoretical probability with examples.
Think about how each type of probability is calculated and the basis for their calculations.
Solution
Empirical probability is based on actual experiments and observations, like flipping a coin 100 times and observing 55 heads, giving an empirical probability of heads as 55/100. Theoretical probability is based on expected outcomes under ideal conditions, like the theoretical probability of heads in a fair coin toss is 1/2. The key difference is empirical probability relies on actual data, while theoretical probability relies on mathematical models.
A die is rolled twice. What is the probability that the sum of the numbers on the two rolls is 7?
List all possible pairs that add up to 7 and count them.
Solution
There are 6 possible outcomes for each die roll, making 36 total possible outcomes when rolling two dice. The pairs that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), totaling 6 favorable outcomes. So, the probability is 6/36 = 1/6.
Explain why the probability of an impossible event is 0 and a sure event is 1.
Consider the definitions of impossible and sure events in terms of favorable outcomes.
Solution
An impossible event has no favorable outcomes out of all possible outcomes, so its probability is 0. A sure event includes all possible outcomes as favorable, so its probability is 1. For example, the probability of rolling a 7 on a standard die is 0 (impossible), and the probability of rolling a number between 1 and 6 is 1 (sure).
A card is drawn from a well-shuffled deck of 52 cards. What is the probability that it is either a king or a heart?
Remember to subtract the overlap (king of hearts) to avoid double-counting.
Solution
There are 4 kings and 13 hearts in a deck, but the king of hearts is counted in both. So, the number of favorable outcomes is 4 (kings) + 13 (hearts) - 1 (king of hearts) = 16. The total number of cards is 52. Thus, the probability is 16/52 = 4/13.
Two coins are tossed simultaneously. What is the probability of getting at least one head?
List all possible outcomes and count those that meet the condition.
Solution
The possible outcomes are HH, HT, TH, TT. The outcomes with at least one head are HH, HT, TH, totaling 3 favorable outcomes out of 4 possible. So, the probability is 3/4.
A box contains 10 bulbs, 4 of which are defective. If 3 bulbs are drawn at random without replacement, what is the probability that none are defective?
Calculate the probability step by step, adjusting the number of bulbs left after each draw.
Solution
First, calculate the number of non-defective bulbs: 10 total - 4 defective = 6 non-defective. The probability of drawing the first non-defective bulb is 6/10, the second is 5/9, and the third is 4/8. The combined probability is (6/10)*(5/9)*(4/8) = 120/720 = 1/6.
What is the probability that a leap year selected at random will have 53 Sundays?
Focus on the extra days beyond the 52 weeks in a leap year.
Solution
A leap year has 366 days, which is 52 weeks and 2 extra days. The extra days can be any of the 7 possible pairs of consecutive days. For there to be 53 Sundays, one of the extra days must be a Sunday. There are 2 favorable pairs: (Sunday, Monday) and (Saturday, Sunday). So, the probability is 2/7.
A number is chosen at random from 1 to 50. What is the probability that it is a prime number?
Recall the definition of a prime number and count how many exist in the given range.
Solution
First, list the prime numbers between 1 and 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, totaling 15 primes. The total numbers are 50. So, the probability is 15/50 = 3/10.
In a class of 40 students, 25 are girls. If a student is chosen at random, what is the probability that the student is a boy?
Subtract the number of girls from the total to find the number of boys.
Solution
First, find the number of boys: 40 total - 25 girls = 15 boys. The probability of choosing a boy is 15/40 = 3/8.
Probability - Challenge Worksheet
Push your limits with complex, exam-level long-form questions.
The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Probability in Class X.
Advanced critical thinking
Test your mastery with complex questions that require critical analysis and reflection.
Questions
A bag contains 5 red, 3 blue, and 2 green marbles. If two marbles are drawn at random without replacement, what is the probability that both marbles are of the same color?
Consider the total number of marbles and how drawing one affects the next draw.
Solution
Calculate the probability for each color separately and then add them together. For red: (5/10)*(4/9), for blue: (3/10)*(2/9), for green: (2/10)*(1/9). Sum these probabilities for the final answer.
In a game, a player rolls two dice. If the sum is 7 or 11, the player wins. What is the probability of winning on the first roll?
Enumerate all possible pairs that sum to 7 or 11.
Solution
List all possible outcomes (36) and count the favorable ones for sums 7 and 11. There are 6 ways to get 7 and 2 ways to get 11, so total favorable outcomes are 8.
A card is drawn from a deck of 52 cards. What is the probability that it is either a heart or a queen?
Remember to subtract the probability of the intersection of the two events.
Solution
Use the formula P(A or B) = P(A) + P(B) - P(A and B). Here, P(heart) = 13/52, P(queen) = 4/52, P(heart and queen) = 1/52.
Two coins are tossed simultaneously. What is the probability of getting at least one head?
It's easier to consider the opposite of 'at least one head'.
Solution
Calculate the probability of the complementary event (no heads, i.e., two tails) and subtract from 1. P(at least one head) = 1 - P(two tails) = 1 - (1/4).
A die is rolled twice. What is the probability that the sum of the numbers is divisible by 3?
Consider all pairs (1,2), (2,1), etc., that sum to multiples of 3.
Solution
List all possible outcomes (36) and count those where the sum is 3, 6, 9, or 12. There are 12 such outcomes.
In a lottery, a ticket has 5 different numbers from 1 to 50. What is the probability of winning if the ticket matches all 5 numbers drawn?
This is a combination problem where order does not matter.
Solution
The probability is 1 divided by the number of combinations of 50 things taken 5 at a time, which is 1/C(50,5).
A box contains 10 bulbs, 4 of which are defective. If 3 bulbs are drawn at random, what is the probability that exactly one is defective?
Think in terms of combinations for selecting defective and non-defective bulbs.
Solution
Use combinations: C(4,1)*C(6,2)/C(10,3). This calculates the number of ways to choose 1 defective and 2 good bulbs over the total ways to choose any 3 bulbs.
What is the probability that a randomly chosen two-digit number is divisible by both 2 and 5?
Numbers divisible by both 2 and 5 are divisible by their LCM, which is 10.
Solution
A number divisible by both 2 and 5 must be divisible by 10. Count the two-digit numbers divisible by 10 (from 10 to 90) and divide by the total two-digit numbers (90-9=81).
A family has two children. Given that at least one is a boy, what is the probability that both are boys?
List all possible gender combinations and exclude the ones that don't meet the condition.
Solution
The possible gender combinations are BB, BG, GB, GG. Given at least one boy, we exclude GG. So, probability of BB is 1/3.
A number is chosen at random from the first 100 natural numbers. What is the probability that it is a perfect square but not a perfect cube?
First, find how many numbers are perfect squares, then subtract those that are both squares and cubes.
Solution
Count the perfect squares (10) and subtract those that are also perfect cubes (1 and 64). So, 10-2=8. Probability is 8/100.
Explore real-world applications of trigonometry in measuring heights, distances, and angles in various fields such as astronomy, navigation, and architecture.
Explore the properties, theorems, and applications of circles in geometry, including tangents, chords, and angles subtended by arcs.
Explore the concepts of calculating areas related to circles, including sectors, segments, and combinations with other geometric shapes.
Explore the concepts of calculating surface areas and volumes of various geometric shapes, including cubes, cylinders, cones, and spheres, to solve real-world problems.
Statistics is the chapter that deals with the collection, analysis, interpretation, presentation, and organization of data.
