Question

Joe has three six‐sided dice that he rolls simultaneously. The faces of the dice contain these letters:

Die #1: A B C D E F

Die #2: G H I J K L

Die #3: M N O P Q R


What is the probability that he will roll three vowels? (8 points; 2 per answer)

Die #1 probability: 13 Die #2 probability: 16 Die #3 probability: 16

Multiply the probabilities to get an answer of: 1108



What is the probability that he will roll the word DIP? (8 points; 2 per answer)

Die #1 probability: 16 Die #2 probability: 16 Die #3 probability: 16

Multiply the probabilities to get an answer of: ??



What is the probability that he will roll the word FAR? (8 points; 2 per answer)

Die #1 probability: ?? Die #2 probability: ?? Die #3 probability: ??

Multiply the probabilities to get an answer of: ??

Answer 1

To calculate the probabilities, we first need to examine which letters appear on each die and identify which letters correspond to vowels or the letters required for the words "DIP" and "FAR."

### Die Faces:
- **Die #1**: A, B, C, D, E, F (vowels: A, E)
- **Die #2**: G, H, I, J, K, L (vowels: I)
- **Die #3**: M, N, O, P, Q, R (vowels: O)

### 1. Probability of rolling three vowels

Vowels on the dice:
- Die #1: 2 vowels (A, E)
- Die #2: 1 vowel (I)
- Die #3: 1 vowel (O)

The total number of outcomes when rolling three dice is \(6 \times 6 \times 6 = 216\).

- Probability from **Die #1** (rolling a vowel):
\[
P(\text{Vowel from Die #1}) = \frac{2}{6} = \frac{1}{3}
\]

- Probability from **Die #2** (rolling a vowel):
\[
P(\text{Vowel from Die #2}) = \frac{1}{6}
\]

- Probability from **Die #3** (rolling a vowel):
\[
P(\text{Vowel from Die #3}) = \frac{1}{6}
\]

Now multiply the probabilities:
\[
P(\text{Three vowels}) = \left(\frac{1}{3}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{3} \times \frac{1}{36} = \frac{1}{108}
\]

### 2. Probability of rolling the word DIP

- **DIP** requires:
- **D** from Die #1 (1/6 since only 1 face has D)
- **I** from Die #2 (1/6 since only 1 face has I)
- **P** from Die #3 (1/6 since only 1 face has P)

Calculating the probabilities:
\[
P(D) = \frac{1}{6}, \quad P(I) = \frac{1}{6}, \quad P(P) = \frac{1}{6}
\]

Now multiply the probabilities:
\[
P(DIP) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}
\]

### 3. Probability of rolling the word FAR

- **FAR** requires:
- **F** from Die #1 (1/6 since only 1 face has F)
- **A** from Die #2 (not possible since A is not on Die #2; should be 0)
- **R** from Die #3 (1/6 since only 1 face has R)

However, A is not on Die #2, so this combination cannot occur.

Thus, the probability \(P(FAR)\) = 0.

### Summary of Results:

- Probability of rolling three vowels: \(\frac{1}{108}\)
- Probability of rolling the word DIP: \(\frac{1}{216}\)
- Probability of rolling the word FAR: 0

### Final Answers:
- Die #1 probability for "DIP": 1/6
- Die #2 probability for "DIP": 1/6
- Die #3 probability for "DIP": 1/6
- Multiply the probabilities to get an answer of: \(\frac{1}{216}\)

- Die #1 probability for "FAR": 1/6 (F)
- Die #2 probability for "FAR": 0 (A is not present)
- Die #3 probability for "FAR": 1/6 (R)
- Multiply the probabilities to get an answer of: 0