
Step-by-step explanation:
Probability=
Probability for a randomly chosen girl to be senior=
Probability for a randomly chosen girl to be senior=
Probability for a randomly chosen boy to be senior=
Probability for a randomly chosen girl to be senior=
For two independent events,
Probability for both event 1 and event 2 to take place=
Since choosing boys and girls is independent,
Probability for both boy an girl chosen to be senior=
Probability for both boy and girl chosen to be senior=
So,required probability is 
A = {1, 3, 5, 7, 9} B = {2, 4, 6, 8, 10} C = {1, 5, 6, 7, 9} A ∩ (B ∪ C) =
vovikov84 [41]
A = {1, 3, 5, 7, 9}
B = {2, 4, 6, 8, 10}
C = {1, 5, 6, 7, 9}
(B ∪ C) = {1, 2, 4, 5, 6, 7, 8, 9, 10}
so
A ∩ (B ∪ C) = {1, 5, 7 , 9}
Answer:
4 14/15
Step-by-step explanation:
8 3/5 - 11/3
8 9/15 - 55/15
8 9/15 - 3 10/15
7 24/15 - 3 10/15
4 14/15