To get the probability of an event to occur, we have the following formula:
[tex]P=\frac{no.\text{ of favorable outcomes}}{\text{total no. of possible outcomes}}[/tex]According to the problem, the sample space is (1, 6, 9, 11, 12, 14, 17, 18, 19, 20) therefore, the total no. of possible outcomes is 10.
For Set B, the sample is (9, 12, 14, 17, 18), therefore, there are 5 possible outcomes that belong to set B.
Starting with the first question, what is the probability of Set B to occur?
[tex]P=\frac{no.\text{ of outcomes from B}}{\text{total no. of possible outcomes}}=\frac{5}{10}=\frac{1}{2}=0.50=50\text{ percent}[/tex]For Set C, the sample is (6,9,11, 12, 18, 19) therefore, there are 5 possible outcomes that belong to set C as well.
On the next question, what is the probability of Set C to occur?
[tex]P=\frac{no.\text{ of outcomes from C}}{\text{total no. of possible outcomes}}=\frac{5}{10}=\frac{1}{2}=0.50=50\text{ percent}[/tex]For the third question, what is the probability of Set B or C to occur?
Since the outcomes under B or C are (6, 9, 11, 12, 14, 17, 18, 19), the probability of the union of B and C is:
[tex]P=\frac{no.\text{ of outcomes from B or C}}{\text{total no. of possible outcomes}}=\frac{8}{10}=\frac{4}{5}=0.80=80\text{ percent}[/tex]On to the last question, what is the probability of the intersection of B and C to occur?
Since the outcomes that are found on both B and C are (9,12,18), the probability of the intersection of B and C is:
[tex]P=\frac{no.\text{ of outcomes found on both B and C}}{\text{total no. of possible outcomes}}=\frac{3}{10}=0.30=30\text{ percent}[/tex]