Answer :
## Question 1
### 1.1 A fair die is rolled once.
#### 1.1.1 Write down the sample space.
The sample space for rolling a fair six-sided die once includes all the possible outcomes. These outcomes are:
[tex]\[ \{1, 2, 3, 4, 5, 6\} \][/tex]
#### 1.1.2 Determine the probability of getting the following:
1. The number 6:
- The probability of rolling a 6 is the ratio of the favorable outcomes (getting a 6) to the total outcomes.
- There is 1 favorable outcome (rolling a 6) and 6 possible outcomes in the sample space.
- [tex]\[ P(\text{6}) = \frac{1}{6} \][/tex]
- Numerical answer: [tex]\( 0.16666666666666666 \)[/tex]
2. A number less than 3:
- Numbers less than 3 on a die are 1 and 2. Hence, there are 2 favorable outcomes.
- [tex]\[ P(\text{less than 3}) = \frac{2}{6} = \frac{1}{3} \][/tex]
- Numerical answer: [tex]\( 0.3333333333333333 \)[/tex]
3. The number 3 or 6:
- The favorable outcomes here are rolling a 3 or a 6. Hence, there are 2 favorable outcomes.
- [tex]\[ P(\text{3 or 6}) = \frac{2}{6} = \frac{1}{3} \][/tex]
- Numerical answer: [tex]\( 0.3333333333333333 \)[/tex]
4. The number 3 and 6:
- It is impossible to roll both a 3 and a 6 simultaneously with one die.
- Hence, there are 0 favorable outcomes.
- [tex]\[ P(\text{3 and 6}) = \frac{0}{6} = 0 \][/tex]
- Numerical answer: [tex]\( 0.0 \)[/tex]
### 1.2 A bag contains 7 red marbles and 5 green marbles. One marble is drawn out of the bag.
#### 1.2.1 Determine the probability that it is a red marble.
- There are 7 red marbles out of a total of 12 marbles (7 red + 5 green).
- [tex]\[ P(\text{red marble}) = \frac{7}{12} \][/tex]
- Numerical answer: [tex]\( 0.5833333333333334 \)[/tex]
#### 1.2.2 Determine the probability that it is a green marble.
- There are 5 green marbles out of a total of 12 marbles.
- [tex]\[ P(\text{green marble}) = \frac{5}{12} \][/tex]
- Numerical answer: [tex]\( 0.4166666666666667 \)[/tex]
#### 1.2.3 Determine the probability that it is a yellow marble.
- There are no yellow marbles in the bag, so the number of favorable outcomes is 0.
- [tex]\[ P(\text{yellow marble}) = \frac{0}{12} = 0 \][/tex]
- Numerical answer: [tex]\( 0.0 \)[/tex]
#### 1.2.4 Determine the probability that it is a red or green marble.
- The events of drawing a red marble and drawing a green marble are mutually exclusive.
- The sum of the probabilities of drawing a red marble or a green marble:
- [tex]\[ P(\text{red or green marble}) = P(\text{red marble}) + P(\text{green marble}) = \frac{7}{12} + \frac{5}{12} = 1 \][/tex]
- Numerical answer: [tex]\( 1.0 \)[/tex]
#### 1.2.5 Determine the probability that it is not a red marble.
- The probability of not drawing a red marble is the complement of the probability of drawing a red marble.
- [tex]\[ P(\text{not red marble}) = 1 - P(\text{red marble}) = 1 - \frac{7}{12} = \frac{5}{12} \][/tex]
- Numerical answer: [tex]\( 0.41666666666666663 \)[/tex]
### 1.1 A fair die is rolled once.
#### 1.1.1 Write down the sample space.
The sample space for rolling a fair six-sided die once includes all the possible outcomes. These outcomes are:
[tex]\[ \{1, 2, 3, 4, 5, 6\} \][/tex]
#### 1.1.2 Determine the probability of getting the following:
1. The number 6:
- The probability of rolling a 6 is the ratio of the favorable outcomes (getting a 6) to the total outcomes.
- There is 1 favorable outcome (rolling a 6) and 6 possible outcomes in the sample space.
- [tex]\[ P(\text{6}) = \frac{1}{6} \][/tex]
- Numerical answer: [tex]\( 0.16666666666666666 \)[/tex]
2. A number less than 3:
- Numbers less than 3 on a die are 1 and 2. Hence, there are 2 favorable outcomes.
- [tex]\[ P(\text{less than 3}) = \frac{2}{6} = \frac{1}{3} \][/tex]
- Numerical answer: [tex]\( 0.3333333333333333 \)[/tex]
3. The number 3 or 6:
- The favorable outcomes here are rolling a 3 or a 6. Hence, there are 2 favorable outcomes.
- [tex]\[ P(\text{3 or 6}) = \frac{2}{6} = \frac{1}{3} \][/tex]
- Numerical answer: [tex]\( 0.3333333333333333 \)[/tex]
4. The number 3 and 6:
- It is impossible to roll both a 3 and a 6 simultaneously with one die.
- Hence, there are 0 favorable outcomes.
- [tex]\[ P(\text{3 and 6}) = \frac{0}{6} = 0 \][/tex]
- Numerical answer: [tex]\( 0.0 \)[/tex]
### 1.2 A bag contains 7 red marbles and 5 green marbles. One marble is drawn out of the bag.
#### 1.2.1 Determine the probability that it is a red marble.
- There are 7 red marbles out of a total of 12 marbles (7 red + 5 green).
- [tex]\[ P(\text{red marble}) = \frac{7}{12} \][/tex]
- Numerical answer: [tex]\( 0.5833333333333334 \)[/tex]
#### 1.2.2 Determine the probability that it is a green marble.
- There are 5 green marbles out of a total of 12 marbles.
- [tex]\[ P(\text{green marble}) = \frac{5}{12} \][/tex]
- Numerical answer: [tex]\( 0.4166666666666667 \)[/tex]
#### 1.2.3 Determine the probability that it is a yellow marble.
- There are no yellow marbles in the bag, so the number of favorable outcomes is 0.
- [tex]\[ P(\text{yellow marble}) = \frac{0}{12} = 0 \][/tex]
- Numerical answer: [tex]\( 0.0 \)[/tex]
#### 1.2.4 Determine the probability that it is a red or green marble.
- The events of drawing a red marble and drawing a green marble are mutually exclusive.
- The sum of the probabilities of drawing a red marble or a green marble:
- [tex]\[ P(\text{red or green marble}) = P(\text{red marble}) + P(\text{green marble}) = \frac{7}{12} + \frac{5}{12} = 1 \][/tex]
- Numerical answer: [tex]\( 1.0 \)[/tex]
#### 1.2.5 Determine the probability that it is not a red marble.
- The probability of not drawing a red marble is the complement of the probability of drawing a red marble.
- [tex]\[ P(\text{not red marble}) = 1 - P(\text{red marble}) = 1 - \frac{7}{12} = \frac{5}{12} \][/tex]
- Numerical answer: [tex]\( 0.41666666666666663 \)[/tex]
