Marcos is selecting classes for next year. He plans to take English, physics, government, pre-calculus, Spanish, and journalism. His school has a six-period day, so he will have one of these classes each period.
1. How many different schedules are possible?
2. How many schedules are possible with first-period pre-calculus?
3. What is the probability that Marcos will get first-period pre-calculus?
4. What is the probability that Marcos will get both first-period pre-calculus and second-period physics?
1. 720
1. Well, if you think about it, Marcos has 6 options for his first class, then 5 options for his second class, 4 options for his third class, and so on. So the total number of different schedules is 6 x 5 x 4 x 3 x 2 x 1, which is 720.
2. If Marcos is doomed to start his day with pre-calculus, then he has already made one decision. So now he has 5 options for his second class, 4 options for his third class, and so on. Therefore, the number of schedules with first-period pre-calculus is 5 x 4 x 3 x 2 x 1, which is 120.
3. The probability of getting first-period pre-calculus is simply the number of schedules with first-period pre-calculus (which is 120) divided by the total number of possible schedules (which is 720). So the probability would be 120/720, which simplifies to 1/6.
4. Ah, now we are getting into some probability mischief. To calculate this probability, we need to figure out how many schedules satisfy both conditions: having first-period pre-calculus and second-period physics. Lucky for us, we already know the number of schedules with first-period pre-calculus (120). Now, since Marcos has already chosen his first class, he only has 4 options left for his second class. After that, he has 3 options for his third class, and so on. Therefore, the number of schedules with first-period pre-calculus and second-period physics is 4 x 3 x 2 x 1, which is 24. So the probability would be 24/720, which simplifies to 1/30.
Hope that brings a smile to your face, even if the probability of getting both classes might not be as high as you'd like!
1. To determine the number of different schedules possible, we use the concept of permutations. Marcos has 6 classes to choose from, and each period can have one class. Therefore, the number of different schedules is given by 6!, which is equal to 720.
2. If Marcos has first-period pre-calculus, then he has already chosen one class, and the remaining periods must be filled with the remaining 5 classes. So there are 5! different schedules possible with first-period pre-calculus, which is equal to 120.
3. The probability of getting first-period pre-calculus is the number of schedules with pre-calculus in the first period divided by the total number of schedules. So the probability is 120/720, which simplifies to 1/6.
4. To determine the probability of getting both first-period pre-calculus and second-period physics, we consider two events. The first event is getting first-period pre-calculus, which has a probability of 1/6. The second event is getting second-period physics, given that Marcos already has pre-calculus in the first period. Since Marcos has 5 classes remaining to fill the remaining 5 periods, the probability of getting physics in the second period is 1/5.
To calculate the probability of both events occurring, we multiply their probabilities together. Therefore, the probability of getting both first-period pre-calculus and second-period physics is (1/6) * (1/5), which simplifies to 1/30.
To answer these questions, we can use the fundamental principle of counting and probability. Let's break down each question step by step:
1. How many different schedules are possible?
To determine the number of different schedules, we need to calculate the number of ways Marcos can arrange his classes. Since there are 6 periods in a day and 6 classes, we can use the concept of permutations to find the total number of arrangements. The formula for permutations is nPr = n! / (n-r)!, where n is the total number of items and r is the number of items to be chosen. In this case, n = 6, and r = 6 since Marcos has to take each class. Therefore, the number of different schedules is 6!.
2. How many schedules are possible with first-period pre-calculus?
To calculate the number of schedules with first-period pre-calculus, we need to fix pre-calculus in the first period and then arrange the remaining 5 classes in the remaining 5 periods. So the number of different schedules with first-period pre-calculus is (5!)*(1!), where 5! is the number of ways to arrange the other 5 classes and 1! represents the fixed pre-calculus class.
3. What is the probability that Marcos will get first-period pre-calculus?
The probability of getting first-period pre-calculus is equal to the number of schedules with first-period pre-calculus divided by the total number of different schedules. So we divide the answer from question 2 by the answer from question 1.
4. What is the probability that Marcos will get both first-period pre-calculus and second-period physics?
To calculate the probability of getting both first-period pre-calculus and second-period physics, we need to fix both classes in the respective periods and arrange the remaining 4 classes in the remaining 4 periods. So the number of schedules with both pre-calculus and physics fixed in the correct periods is (4!)*(1!)*(1!), with 4! representing the number of ways to arrange the remaining 4 classes, 1! representing the fixed pre-calculus class, and 1! representing the fixed physics class. To calculate the probability, we divide this number by the total number of different schedules (the answer from question 1).
Now you have the explanation on how to find the answers to these questions. Plug in the values into the formulas explained above, and you'll get the specific numerical answers.