Day 2 of this challenge series comprises three parts/challenges that focus on the concepts of Dice, Probability, and Compound Event Probability.
This challenge series does not involve any code input; instead, it is presented in multiple-choice question (MCQ) format, making it easily approachable.
To ensure a solid foundation, let’s begin by clarifying the fundamental concept of Probability before delving into the challenges.
Probability -
- In simple terms, probability is like a way of predicting the future based on what we know about the past or the current situation. It’s like making an educated guess about what might happen next.
- For example, if you toss a fair coin, there are two possible outcomes: heads or tails. The probability of getting heads is 1 out of 2 (or 1/2), because there is one chance out of two equally likely outcomes.
- Probability helps us make decisions and understand the likelihood of different outcomes. It is used in many areas of life, such as predicting weather conditions, analyzing risks in games of chance, or determining the chances of success in various situations.
- Remember, probability is not a guarantee of what will happen, but rather an estimation of what is likely to happen based on the available information.
Probability = Number of favorable outcomes / Total number of outcomes basically n(A)/n(S)
Sample space -
- The sample space is like a list or collection of all the different things that could happen when an experiment or event takes place.
- For example, if you roll a fair six-sided die, the sample space would consist of all the possible outcomes, which are the numbers 1, 2, 3, 4, 5, and 6. So, in this case, the sample space would be {1, 2, 3, 4, 5, 6}.
Event -
- An event is a particular situation or occurrence that we want to study or assign a probability to. It can be a single outcome or a combination of outcomes from the sample space.
- For example, if we roll a fair six-sided die, some possible events could be: Getting an even number: {2, 4, 6}, Rolling a 1 or a 6: {1, 6},Rolling a number greater than 3: {4, 5, 6}
Compound Event Probability-
- Compound event probability refers to the chance of multiple events happening together. The way we calculate this probability depends on whether the events are independent or dependent.
- Independent Events: If two events are independent, it means that one event happening does not affect the other event. To find the probability of two independent events happening together, you multiply their individual probabilities.
- Dependent Events: If two events are dependent, it means that the outcome of one event affects the outcome of the other event. To find the probability of two dependent events happening together, you multiply the probability of the first event by the conditional probability of the second event given that the first event has already happened.
Let’s begin with the first challenge.
Challenge #1 — Basic Probability
Question -
- In a single toss of 2 fair (evenly-weighted) six-sided dice, find the probability that their sum will be at most 9.
Answer — 5/6
Explanation -
- The question is about tossing two dice and finding the probability that their sum will be 9.
- To start, we need to create the sample space, which represents all possible outcomes when two dice are tossed. The sample space is:
- 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)}
- The sample space contains 36 possible outcomes. n(S) = 36
- Now, let’s define the event A as the sum of the two dice being at most 9. The outcomes in event A are:
- A = {(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), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3)}
- The event A consists of 30 outcomes. n(A) = 30
- To calculate the probability of event A, we divide the number of outcomes in event A by the total number of outcomes in the sample space:
- P(A) = n(A) / n(S) = 30 / 36 = 5 / 6
- Therefore, the probability that the sum of the two dice will be at most 9 is 5/6.
Challenge #2 — More Dice
Question -
- In a single toss of fair (evenly-weighted) six-sided dice, find the probability that the values rolled by each die will be different and the two dice have a sum of .
Answer — 1/9
Explanation -
- Lets S be sample space
- 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)}
- The sample space contains 36 possible outcomes.
- n(S) = 36
- In this scenario, the question states that the number on one dice should not be the same as the number on the other dice, and the sum of their numbers should be 6. We have two conditions to fulfill based on the question.
- Let’s define event A as the set of outcomes that satisfy these conditions:
- A = {(1, 5), (2, 4), (4, 2), (5, 1)}
- The set A consists of 4 possible outcomes that meet the given conditions.
- To calculate the probability, we need to divide the number of outcomes in event A by the total number of outcomes in the sample space.
- Considering the sample space S from the previous information (with 36 outcomes), we have:
- P(A) = n(A) / n(S) = 4 / 36 = 1 / 9
- Therefore, the probability of event A occurring is 1/9.
Challenge #3 — Compound Event Probability
Question -
There are 3 urns labeled X , Y ,Z
- Urn X contains 4 red balls and 3 black balls.
- Urn Y contains 5 red balls and 4 black balls.
- Urn Z contains 4 red balls and 4 black balls.
One ball is drawn from each of the 3 urns. What is the probability that, 3 of the balls drawn, 2 are red and 1 is black?
Answer — 17/42
Explanation -
- Sample Space:
- We have three urns: X, Y, and Z, each containing different combinations of red (R) and black (B) balls.
- Urn X: {1R, 2R, 3R, 4R, 3B, 1B, 2B} with a total of 7 balls (n(X) = 7).
- Urn Y: {1R, 2R, 3R, 4R, 5R, 1B, 2B, 3B, 4B} with a total of 9 balls (n(Y) = 9).
- Urn Z: {1R, 2R, 3R, 4R, 1B, 2B, 3B, 4B} with a total of 8 balls (n(Z) = 8).
2. Probability of Red:
- Probability of getting red from each urn:
- Urn X: P(R1) = 4/7 (4 red balls out of 7 total balls).
- Urn Y: P(R2) = 5/9 (5 red balls out of 9 total balls).
- Urn Z: P(R3) = 4/8 (4 red balls out of 8 total balls).
3. Probability of Black:
- Probability of getting black from each urn:
- Urn X: P(B1) = 3/7 (3 black balls out of 7 total balls).
- Urn Y: P(B2) = 4/9 (4 black balls out of 9 total balls).
- Urn Z: P(B3) = 4/8 (4 black balls out of 8 total balls).
4. Probability of Specific Combination:
- The condition is to find the probability of drawing 3 balls, consisting of 2 red and 1 black, one ball from each urn.
- Let’s assume the event E represents this condition.
- The probability of event E can be calculated as follows:
- P(E) = P(R1) * P(R2) * P(B3) + P(R1) * P(B2) * P(R3) + P(B1) * P(R2) * P(R3)
- Substitute the values: P(E) = (4/7) * (5/9) * (4/8) + (4/7) * (4/9) * (4/8) + (3/7) * (5/9) * (4/8)
- Calculate to get the result: P(E) = 17/42
Therefore, the probability of drawing 3 balls with 2 red and 1 black, one ball from each urn, is 17/42.
Exciting blogs are on the way! Stay connected by following me for more engaging content.
