SOLUTION: can someone please help me with this problem: Among 170 households surveyed, 43 have a video camera, 44 have a snapshot camera, 48 binoculars, 7 have a video camera and a snapsh

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Question 162955: can someone please help me with this problem:
Among 170 households surveyed, 43 have a video camera, 44 have a snapshot camera, 48 binoculars, 7 have a video camera and a snapshot camera, 8 have a snapshot camera and binoculars, 4 have all three products. What is the probability that a household will have a snapshot camera or binoculars? Express the answer as a fraction.
a. 79/170
b. 46/85
c. 42/85
d. 87/170
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
                      44
P(snapshot camera)= ------ 
                     170


                 48
P(binoculars)= ------ 
                170



                                     8
P(snapshot camera AND binoculars)= ------ 
                                    170


To find the probability of either A or B (but not both), simply add the individual probabilities of A and B and subtract off the probability of A and B like this:

P(A or B) = P(A)+P(B)-P(A and B)

So in this case: 


P(snapshot camera OR binoculars) = P(snapshot camera) + P(binoculars) - P(snapshot camera AND binoculars)


                                     44       48        8        84      42
P(snapshot camera OR binoculars) = ------ + ------ -  ----- =  ------ = ----
                                    170      170       170      170      85



So the probability of a person having a snapshot camera OR binoculars (but not both) is 42%2F85 which means that the answer is c)