Sure, let's break down each part of the problem and solve it step by step.
### Step 1: Distribution of Cards
Given:
- Total baseball cards: 0 (before giving away any cards)
- Total football cards: 0 (before giving away any cards)
- Total basketball cards: 0 (before giving away any cards)
He gives his friend:
- 8 baseball cards
- 10 football cards
- 7 basketball cards
Thus, he has:
- 0 - 8 baseball cards = -8 (this does not make sense in the context, as he cannot give away cards he doesn't have)
- 0 - 10 football cards = -10 (similarly, this does not make sense)
- 0 - 7 basketball cards = -7 (again, this does not make sense)
Given that the total cards he had in the problem initially was 0 for each type, a correction might be needed here. Let's check each card type he could give:
Corrected total remaining cards:
- Remaining baseball cards: 0
- Remaining football cards: 0
- Remaining basketball cards: 0
This indicates he cannot give away more cards than he initially had.
### Step 2: Probability using a 12-sided die
A 12-sided solid (dice) is numbered from 1 to 12.
To find the probabilities:
1. P(number greater than 10):
- Numbers greater than 10 are 11 and 12. There are 2 favorable outcomes.
- Total possible outcomes = 12.
- Probability = Number of favorable outcomes / Total possible outcomes = 2/12 = 1/6.
2. P(number less than 5):
- Numbers less than 5 are 1, 2, 3, and 4. There are 4 favorable outcomes.
- Total possible outcomes = 12.
- Probability = Number of favorable outcomes / Total possible outcomes = 4/12 = 1/3.
### Step 3: Rolling the 12-sided die 200 times
Finding the number of times a specific set of numbers would appear in 200 rolls:
To find the expected frequency of 4, 6, or 9:
- There are 3 numbers: 4, 6, and 9.
- Probability of rolling a 4, 6, or 9 = 3 / 12 = 1/4.
Expected frequency in 200 rolls:
- Expected number of times = (1/4) 200 = 50.
### Step 4: Rolling a single number cube (6-sided die) twice
Analysis of the question:
If we're considering a single 6-sided die, then rolling it twice produces an even larger set of outcomes. Though the question isn't completely clear, if we were to analyze:
Rolling the die twice:
- Total outcomes when rolling a die twice = 6 6 = 36.
- Each side is equally likely to appear in any roll out of these 36.
### Summary
1. Distribution of Cards:
- Remaining baseball cards: 0
- Remaining football cards: 0
- Remaining basketball cards: 0
2. Probability Findings:
- P(number greater than 10) = 1/6.
- P(number less than 5) = 1/3.
3. Expected frequency of 4, 6, or 9 in 200 rolls:
- Expected times either 4, 6, or 9 is rolled: 50.
I hope this provides a clear step-by-step solution to each part of the given problem!
