SOLUTION: A survey of 448 students at a small college showed that 234 students responded that they did not like to ski, 100 students responded that they played hockey, and that there were

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Question 1138803: A survey of 448 students at a small college showed that 234 students responded
that they did not like to ski, 100 students responded that they played hockey,
and that there were 189 students the neither skied nor played hockey. Without
the use of a Venn Diagram, determine the probability that a randomly selected
student both skied and played hockey. Make sure that you define all the
appropriate sets needed to solve the problem, convert the phrases in the above
sentences to set theoretic notation, and show the steps.
Answer by Edwin McCravy(20077) About Me  (Show Source):
You can put this solution on YOUR website!
A survey of 448 students at a small college showed that 234 students responded
that they did not like to ski, 100 students responded that they played hockey,
and that there were 189 students the neither skied nor played hockey. Without
the use of a Venn Diagram, determine the probability that a randomly selected
student both skied and played hockey. Make sure that you define all the
appropriate sets needed to solve the problem, convert the phrases in the above
sentences to set theoretic notation, and show the steps. 
Even though you are not to use a Venn diagram, use one privately anyway to get
the idea right and then set up the equations. Don't turn in the Venn diagram.



Let a = n(SᑎH')
Let b = n(SᑎH)
Let c = n(S'ᑎH)
Let d = n(S'ᑎH')

>>A survey of 448 students at a small college...<<

a+b+c+d = 448 = n(U) = n(SᑎH') + n(SᑎH) + n(S'ᑎH) + n(S'ᑎH')

>>...showed that 234 students responded that they did not like to ski,...<<

c+d = 234 = n(S'ᑎH) + n(S'ᑎH')

>>100 students responded that they played hockey,<< 

b+c = 100 =n(SᑎH) + n(S'ᑎH) 

>>and that there were 189 students the neither skied nor played hockey.<< 

d = 189 = n(S'ᑎH')

So we have the system of equations:

system%28a%2Bb%2Bc%2Bd+=+448%2C%0D%0Ac%2Bd+=+234%2C+%0D%0Ab%2Bc+=+100%2C%0D%0Ad+=+189%29

Solve that system by substitution (that's easy), getting 
a=159, b=55, c=45, d=189



Then the desired probability is 

b/(a+b+c+d) = n(SᑎH)/n(U) = 55/448

Edwin