Question 1118900: Can someone tell me if I did this math problem, correctly? Thanks. My answers are:
a. Skippy = 26/114 = 13/57.
b. Peter Pan chunky butter = 51/114 = 17/38.
c. Jif Chunky = 22/114 = 11/57 or Skippy smooth peanut butter = 17/114.
Using the chart below, if Andrew selects one jar at random to place on the shelf, determine the probability she will select a jar of:
a. Skippy
b. Peter Pan chunky butter
c. Jif Chunky or Skippy smooth peanut butter.
Brand Smooth Chunky Total
Peter Pan 24 13 37
Jif 13 22 35
Skippy 17 9 26
Other 9 7 16
Total 63 51 114
Answer by greenestamps(13216)   (Show Source):
You can put this solution on YOUR website!
I think you know what you are doing here; but it seems you aren't always reading or interpreting the questions correctly.
Your answer for part a is fine.
Your answer for part b is the probability for any chunky peanut butter; the question asked for the probability of Skippy peanut butter.
You gave two separate answers for part c. When the question asks for the probability of Jif Chunky OR Skippy smooth, the answer is a single fraction which is the sum of the two answers you show.
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correction....
part b asks for the probability of Peter Pan chunky, not Skippy chunky.
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response to student's further questions....
For part b, your answer was 51/114. 114 is the total; 51 is the total chunky; so your answer is the probability that she chooses chunky peanut butter of ANY brand. The problem asks for the probability that she chooses Peter Pan chunky. Only 13 of the 51 chunky peanut butters are Peter Pan. The answer is 13/114, not 51/114.
For part c, the problem asks for the probability that she chooses EITHER JIf chunky OR Skippy smooth. When a probability question is asked that way, the answer is not the two separate probabilities for each outcome; it is a single answer that is the SUM of those two probabilities. So yes, add the two answers you originally gave to get the single correct answer.
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