Answer:
First figure: An example of set C is {9 , 11 , 13}
Second figure: [tex]P(B')=\frac{6}{11}[/tex]
Step-by-step explanation:
First figure:
Let us explain the meaning of subset
If set X subset set Y, that means every element in set X is in set Y and some elements in set Y are in set X
∵ U is the universal set ⇒ contains all elements in the problem
∵ U = {5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15}
∵ A = {5 , 6 , 7 , 8}
∵ B = {5 , 7 , 9 , 11 , 13 , 5}
- All the elements in set B are odd numbers
∵ C is subset of B
∴ All elements on C belongs to B
- That means the elements in C must be some elements of B
∴ C could be {9 , 11 , 13} OR {5 , 7 , 9 , 15} OR {7 , 9} OR .........
∴ An example of set C is {9 , 11 , 13}
Second figure:
Let us explain the meaning of complement
Complement of a set is all the elements in the universal set except the elements in this set
Let us list U and B from the figure
∵ U = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11}
∵ B = {2 , 3 , 4 , 5 , 7}
- To find B' list all the elements in set U except the elements in set B
∴ B' = {1 , 6 , 8 , 9 , 10 , 11}
To find the probability of B', find n(B') the number of elements of set B' and n(U) the number of elements in set U, then find [tex]P(B')=\frac{n(B')}{n(U)}[/tex]
∵ Set B' has 6 elements
∵ n(B') = 6
∴ Set U has 11 elements
∴ n(U) = 11
∴ [tex]P(B')=\frac{6}{11}[/tex]
