Question 1163614: A hospital spokesperson reported that 4 births had taken place at the Taguig-Pateros Hospital
during the last 24 hrs. Find the following probabilities:
a. P(A) = P(the event that 2 boys and 2 girls are born)
b. P(B) = P(no boys are born)
c. P(C) = P(at least one boy is born)
d. P(A|C)
e. P(B|C)
f. Are A and C mutually exclusive events? Are A and C independent?
g. Are B and C mutually exclusive events? Are B and C independent?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 1--4--6--4--1
2 boys and 2 girls is 6/16 or 0.375 probability
no boys is 1/16 or 0.0625 probability
at least one boy is born is the complement or 0.9375 probability
given at least 1 boy, probability of two boys
this would be 6/15, since given at least one boy decreases denominator from 16 to 15.
P(B|C) is 0.
A and C are not mutually exclusive, B and C are mutually exclusive
P(A)*P(C)= P(A and C)
(5/16)(15/16)=75/256; P(A and C) =P(A)=5/16
They aren't equal so are dependent.
Once you know 2 boys and 2 girls, you know there is at least one boy.
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