Question 1041972: To gather data involving their classmates, 34 fourth graders each filled out a personal information survey sheet. In the category of pets, 16 students had a cat, 21 students had a dog, and 8 students had both a dog and a cat. If a student is randomly selected from this class, what is the probability the student will have a dog or a cat?
Found 2 solutions by jorel555, Edwin McCravy:
Answer by jorel555(1290)   (Show Source):
You can put this solution on YOUR website! 16 students had a cat, 21 had a dog, and 8 had both. So 16+21-8=29 had either a cat or a dog, or both. 29/34=.853 probability that a random student has one or the other, or both. ☺☺☺☺
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website! You can do it like the other tutor told you, but I think
you are supposed to either
1. be learning the formula:
P(A or B) = P(A) + P(B) - P(A and B)
P(cat or dog) = P(cat) + P(dog) - P(cat and dog)
P(cat or dog) = 16/34 + 21/34 - 8/34
P(cat or dog) = 29/34
or
2. learning to use Venn diagrams (sometimes they are
called Euler diagrams). So let's do it that way too:
Everybody within the red circle has a cat.
Everybody within the blue circle has a dog.
Those in the overlapping part of the circles have
both a cat and a dog.
Those outside both circles have neither a cat not a dog.
8 students had both a dog and a cat.
So we put 8 in the middle, the part where the circles
overlap.
16 students had a cat
Those 16 include the 8 that have both a cat and a dog,
as well as those who have only a cat. We already have
put those 8 with both cat and dog in the part of the
red circle that overlaps the blue circle, so to find
out how many have a cat only, we subtract the 8 in the
overlapping part from the 16, getting 8 that have a cat
only. So we put 8 in the left part of the red circle,
to indicate that 8 people have a cat ONLY, and no dog.
21 students had a dog
Those 21 include the 8 that have both a dog and a cat,
as well as those who have only a dog. We already have
put those 8 with both dog and cat in the part of the
red circle that overlaps the blue circle, so to find
out how many have a dog only, we subtract the 8 in the
overlapping part from the 21, getting 13 that have a dog
only. So we put 13 in the right part of the blue circle,
to indicate that 13 people have a dog ONLY, and no cat.
Now we have 8 in the "cat only" category, the left part of the
red circle, 8 in the "cat and dog" category, and 13 in the
"dog only" category. But that's only 8+8+13=29 students that
have either a cat or a dog or both. There are 34 students in
all. So that leaves 34-29=5 students that have neither a cat
nor a dog. So we put 5 outside both circles:
So the probability that a student has either a cat or a dog
is the probability that he is in one of the 8+8+13 = 29 in
the circles, which is 29/34.
Edwin
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