Question 1092185: I teach 7th grade and I am having trouble setting up this problem:
A nationwide survey showed that only 4% of children liked lima beans. What is the probability that any two children will both like lima beans? Book has 1/16 as the answer, but I'm having difficulty setting it up.
Found 3 solutions by Boreal, greenestamps, richwmiller: Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The probability any one child will like lima beans is 0.04. Assuming independence, the probability of another child's liking them is also 0.04. But we do not know the size of the population from which we are choosing. If it is two children, the probability is 0.0016. I wonder if somehow the 1/16 is a misprint.
If it is say 10 children, it is 10C2*0.04^2*0.96^8=0.0519. There are different answers for different numbers of children.
Key question: any two children from what size group?
With some work (or a table), one can show that the probability of EXACTLY two children from a group of 9 will like lima beans is close to 1/16 or 0.0625, although it is not exactly 1/16.
Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website! I interpret the problem differently than the tutor who posted the first response. The way I read the problem, the size of the group is irrelevant.
If the probability of any one child liking lima beans is 1/x, then the probability that two children will both like lima beans is (1/x)^2.
Since the book says the probability of two children both liking lima beans is 1/16, which is (1/4)^2, I'm wondering if the problem said that the probability of any child liking lima beans is 1/4 -- instead of 4%.
If 4% is indeed the probability for each child, then the probability that two children will both like lima beans is (4/100)^2 = 16/10000 = 0.0016 = 0.16%
Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! From my experience as a teacher, the answers in the book are not updated when the problems are updated. So, the book answers are often wrong.
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