SOLUTION: 400 students are surveyed about the pets they have. 250 have a dog, 190 have a cat and 160 have a dog and a cat. What is the probability that a student has a dog? And what is t

There is a universal set U of 400 students.


There is a subset D of 250 students that have a dog.


There is a subset C of 190 students that have a cat.


And there is the intersection (D n C) of these subsets, which contains 160 students that have a dog AND a cat.


Then the number of students who have a dog OR a cat is  n(D U C) = n(D) + n(C) - n(D n C) = 250 + 190 - 160 = 280.


The answer to the first question is  P(D) =  =  = .


The answer to the second question is  P(D U C) =  =  = .

The probability of a dog is 250 out of 400 = 250/400 = 5/8

For the other question I think a Venn diagram is better.

Let D = the set of students who own dogs.
Let C = the set of students who own cats.
Let w = the number of students with dogs but NO cats.
Let x = the number of students both cats and dogs.
Let y = the number of students with cats but NO dogs. 
Let z = the number of students with NO dogs and NO cats.

So w + x + y + z = 400.  

So w + x = 250

So x + y = 190

So x = 160

Substitute 160 for x in w + x = 250
                      w + 160 = 250
                            w = 90

Substitute 160 for x in x + y = 190
                      160 + y = 190
                            y = 30

Substitute w=90, x=160, y=30, in w + x + y + z = 400
                             90 + 160 + 30 + z = 400
                                       280 + z = 400
                                             z = 120

[You weren't asked for z, but you should know how to find it anyway,
in case you'd have been asked about petless students]

That's the number in both circles over 400, or (w+x+y)/400 = (90+160+30)/400 =
280/400 = 7/10

Edwin