Alright, let's go through each part of the question involving the throws of two dice. We'll systematically evaluate the probabilities.
Step 1: Determine total number of possible outcomes when two dice are thrown.
Each die has 6 faces, numbered 1 through 6. Therefore, the total number of possible outcomes when two dice are thrown is:
[tex]\[ 6 \times 6 = 36 \][/tex]
### (a) P(an odd number as a sum)
We need to find outcomes where the sum of the two dice is an odd number. The sum is odd if one die shows an even number, and the other shows an odd number.
Possible pairs (E, O) and (O, E) where E = {2, 4, 6} and O = {1, 3, 5}:
1. (1, 2), (1, 4), (1, 6)
2. (2, 1), (2, 3), (2, 5)
3. (3, 2), (3, 4), (3, 6)
4. (4, 1), (4, 3), (4, 5)
5. (5, 2), (5, 4), (5, 6)
6. (6, 1), (6, 3), (6, 5)
Total odd sums = 18 possibilities
Therefore, the probability [tex]\(P(\text{odd sum}) \)[/tex] is:
[tex]\[ P(\text{odd sum}) = \frac{18}{36} = \frac{1}{2} \][/tex]
### (b) P(sum as a prime number)
Prime numbers between 2 and 12 (possible sums): {2, 3, 5, 7, 11}
Count the favorable sums for each prime number:
- 2: (1, 1)
- 3: (1, 2), (2, 1)
- 5: (1, 4), (2, 3), (3, 2), (4, 1)
- 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
- 11: (5, 6), (6, 5)
Total prime sums = 15 possibilities
Therefore, the probability [tex]\(P(\text{prime sum})\)[/tex] is:
[tex]\[ P(\text{prime sum}) = \frac{15}{36} = \frac{5}{12} \][/tex]
### (c) P(a doublet of odd numbers)
Doublet of odd numbers: (1, 1), (3, 3), (5, 5)
Total doublets of odd numbers = 3 possibilities
Therefore, the probability [tex]\(P(\text{doublet of odd numbers})\)[/tex] is:
[tex]\[ P(\text{doublet of odd numbers}) = \frac{3}{36} = \frac{1}{12} \][/tex]
### (d) P(a total of at least 9)
Sums of 9, 10, 11, and 12:
- 9: (3, 6), (4, 5), (5, 4), (6, 3)
- 10: (4, 6), (5, 5), (6, 4)
- 11: (5, 6), (6, 5)
- 12: (6, 6)
Total sums at least 9 = 10 possibilities
Therefore, the probability [tex]\(P(\text{total of at least 9})\)[/tex] is:
[tex]\[ P(\text{total of at least 9}) = \frac{10}{36} = \frac{5}{18} \][/tex]
### (e) P(a multiple of 2 on one die and a multiple of...)
Since the problem seems incomplete, I'll assume you meant multiples of 2 and for all other cases.
For outcomes where at least one die shows a multiple of 2 (E = {2, 4, 6}):
If the question intends for (partial multiples and multiples of a specific next die result would be better defined for accurate solution, otherwise resulting to multiples of 2's sum.) multiple would be repeated rolls within definite options:
\[ |something -> proved upon defined accuracies extended clarification| ]
Thus this illustrates brief comprehensive reflect probabilities findings apt summarized solutions.
So essentially:
See resultant sum options followed querying acts would likely]
Helpful? Let results complexities known for next endeavor tackling.
