Im confusing the questions with and / or.
table:
Grade level | Bird | Cat | Dog | Fish
6th graders | 9 | 27 | 42 | 3
7th graders | 12 | 36 | 56 | 4
1. what is the probability randomly selected student is an 7th grader and prefers a bird as a pet?
a. 33%
b. 6%
c. 75%
d. 28%
my answer is C, i add all of pets of 7th graders row except bird, (96/108)-(12/108)=77
2. What is the probability randomly selected student is an 7th grader or prefers a bird as a pet?
a. 33%
b. 62%
c. 19%
d. 46%
really struggling with this
I counted 189 students
There are 12 7th graders who like birds
to the prob = 12/189 = .063... or 6.3%
For "and" the event must be in both groups
For "or" it is in one OR the other, including both
but make sure you don't count 7th grader AND birdlover twice
so 7th graders ---- 12+36+56+4 = 108
bird lovers -------- 9+12 = 21
total = 108+21 - 12 = 117
prob = 117/189 = .619 or appr 62%
thank u
To find the probability of a student being a 7th grader and preferring a bird as a pet, you need to find the intersection of the events "student is a 7th grader" and "student prefers a bird as a pet."
In the given table, you can see that 12 out of the 7th graders prefer a bird as a pet. To find the probability, divide the number of 7th graders who prefer a bird (12) by the total number of students (108):
Probability = 12/108 = 1/9 ≈ 0.111
Now, let's move on to the second question, which is asking for the probability of a student being a 7th grader or preferring a bird as a pet. This is an example of the "or" operation.
To find the probability of the union of two events (A or B), you need to add the probabilities of each event separately and then subtract the probability of their intersection, as it was counted twice.
To find the probability of a student being a 7th grader, you can add the number of 7th graders who prefer each type of pet:
12 (bird) + 36 (cat) + 56 (dog) + 4 (fish) = 108.
Next, to find the probability of preferring a bird as a pet, you can add the number of students from each grade level who prefer a bird:
12 (7th graders) + 9 (6th graders) = 21.
However, there is an overlap between these two groups as 12 students are both 7th graders and prefer a bird. Therefore, you need to subtract this intersection from the total count:
Total count = 108 (7th graders) + 21 (bird preference) - 12 (intersection) = 117.
Probability = 117/108 = 1.083 ≈ 0.108.
Therefore, the probability of a randomly selected student being a 7th grader or preferring a bird as a pet is approximately 0.108 or 10.8%.
Based on these calculations, it seems that none of the provided answer choices matches the correct probabilities for the given questions.