Introduction
Welcome to our compound probability worksheet with answers! In this article, we will provide you with a comprehensive set of practice problems to test your understanding of compound probability. Each problem will be accompanied by a detailed solution, allowing you to check your work and learn from any mistakes. Whether you are a student preparing for an exam or simply looking to brush up on your probability skills, this worksheet is designed to help you succeed. Let's dive in!
Problem 1: Independent Events
Problem:
A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. If two marbles are drawn without replacement, what is the probability that both marbles are red?
Solution:
To solve this problem, we first need to determine the total number of marbles in the bag. There are 5 red marbles, 3 blue marbles, and 2 green marbles, so the total number of marbles is 5 + 3 + 2 = 10.
Next, we need to determine the number of ways to choose 2 red marbles from the bag. Since we are drawing without replacement, there will be one less red marble available for the second draw. Therefore, the number of ways to choose 2 red marbles is 5 choose 2, which can be calculated as: 5! / (2! * (5-2)!) = 10.
Finally, we need to determine the total number of ways to choose 2 marbles from the bag. This can be calculated as 10 choose 2, which is: 10! / (2! * (10-2)!) = 45.
Therefore, the probability of drawing two red marbles is 10/45, which simplifies to 2/9.
Problem 2: Dependent Events
Problem:
A bag contains 4 red marbles and 6 blue marbles. If two marbles are drawn with replacement, what is the probability that both marbles are red?
Solution:
In this problem, we are drawing with replacement, which means that after each draw, the marble is put back into the bag. This means that the probability of drawing a red marble on each draw remains the same.
The probability of drawing a red marble on the first draw is 4/10, since there are 4 red marbles and a total of 10 marbles in the bag.
Since we are drawing with replacement, the probability of drawing a red marble on the second draw is also 4/10.
To find the probability of both events occurring, we simply multiply the probabilities together: (4/10) * (4/10) = 16/100, which simplifies to 4/25.
Problem 3: Mutually Exclusive Events
Problem:
A deck of cards contains 52 cards, with 13 cards in each suit (hearts, clubs, diamonds, spades). What is the probability of drawing a heart or a diamond?
Solution:
In this problem, we are dealing with mutually exclusive events, which means that the events cannot occur at the same time. In other words, if we draw a heart, we cannot also draw a diamond (and vice versa).
The probability of drawing a heart is 13/52, since there are 13 hearts in the deck and a total of 52 cards.
The probability of drawing a diamond is also 13/52, since there are 13 diamonds in the deck.
To find the probability of either event occurring, we simply add the probabilities together: 13/52 + 13/52 = 26/52, which simplifies to 1/2.
Problem 4: Conditional Probability
Problem:
A bag contains 8 red marbles and 6 blue marbles. If a marble is drawn at random and is found to be red, what is the probability that the next marble drawn will also be red?
Solution:
In this problem, we are dealing with conditional probability, which means that the probability of an event occurring depends on another event occurring first.
The probability of drawing a red marble on the first draw is 8/14, since there are 8 red marbles and a total of 14 marbles in the bag.
After the first draw, there are now 7 red marbles and 13 marbles in total remaining in the bag.
The probability of drawing a red marble on the second draw, given that the first marble was red, is 7/13.
To find the probability of both events occurring, we simply multiply the probabilities together: (8/14) * (7/13) = 56/182, which simplifies to 8/26.
Problem 5: Complementary Events
Problem:
A fair six-sided die is rolled. What is the probability of rolling a number greater than 4?
Solution:
In this problem, we are dealing with complementary events, which means that the probability of an event occurring is equal to 1 minus the probability of the event not occurring.
The probability of rolling a number greater than 4 is equal to the probability of rolling a 5 or a 6.
Since there are 6 equally likely outcomes (the numbers 1 to 6) and 2 of those outcomes (5 and 6) satisfy the condition, the probability of rolling a number greater than 4 is 2/6, which simplifies to 1/3.
Problem 6: The Multiplication Rule
Problem:
A bag contains 10 red marbles and 8 blue marbles. If two marbles are drawn without replacement, what is the probability that the first marble is red and the second marble is blue?
Solution:
To solve this problem, we can use the multiplication rule, which states that the probability of two independent events occurring is equal to the product of their individual probabilities.
The probability of drawing a red marble on the first draw is 10/18, since there are 10 red marbles and a total of 18 marbles in the bag.
After the first draw, there are now 9 red marbles and 17 marbles in total remaining in the bag.
The probability of drawing a blue marble on the second draw, given that the first marble was red, is 8/17.
To find the probability of both events occurring, we multiply the probabilities together: (10/18) * (8/17) = 80/306, which simplifies to 40/153.
Problem 7: The Addition Rule
Problem:
A bag contains 5 red marbles and 7 blue marbles. If two marbles are drawn without replacement, what is the probability that at least one marble is red?
Solution:
To solve this problem, we can use the addition rule, which states that the probability of two mutually exclusive events occurring is equal to the sum of their individual probabilities.
The probability of drawing a red marble on the first draw is 5/12, since there are 5 red marbles and a total of 12 marbles in the bag.
After the first draw, there are now 4 red marbles and 11 marbles in total remaining in the bag.
The probability of drawing a blue marble on the second draw, given that the first marble was red, is 7/11.
To find the probability of at least one red marble, we need to consider two cases:
Case 1: Red on the first draw, blue on the second draw: (5/12) * (7/11) = 35/132.
Case 2: Blue on the first draw, red on the second draw: (7/12) * (4/11) = 28/132.
Adding these probabilities together, we get 35/132 + 28/132 = 63/132, which simplifies to 21/44.
Problem 8: Permutations
Problem:
A committee of 5 people is being selected from a group of 10 candidates. How many different committees can be formed?
Solution:
To solve this problem, we can use the concept of permutations, which is used to calculate the number of ways to arrange objects in a specific order.
The number of different committees that can be formed is equal to the number of ways to choose 5 people from a group of 10 candidates.
This can be calculated as 10 choose 5, which is equal to: 10! / (5! * (10-5)!) = 10! / (5! * 5!) = 10 * 9 *