Question 787116: The school book store buys calculators in August, of which 8 are defective and 46 are non-defective. A sample of 3 different calculators is randomly selected from this group. What is the probability that all three will be defective?
I need help for my first statistics exam.
Thanks. Leyanes
Answer by xinxin(76)   (Show Source):
You can put this solution on YOUR website! Pr (all three will be defective) = C (8,3)/ C (54,3) = 56/24804 = 0.0023
Which topic do you need help with? I would suggest you review the study guide (or other names your teacher would call). Stats is a difficult subject, to me it's impossible to get a high grade without studying the handouts (study guide, review sheet, or whatever) given by the teacher.
C(8,3) means all 3 defective products are selected from 8 defective products. C(54,3) means randomly select 3 products from all products (8 defective + 46 non-defective= 54); it is the sample space in this case.
C(n,r) comes from combination formula n!/(r!(n-r)!). It is much easier to remember and compute than the original one. Basically C(n,r) = (n*(n-1)*(n-2)*...)/(r*(r-1)*(r-2)*...*1); the number of digits in the numerator is equal to the value of r. For example, C(8,3)=(8*7*6)/(3*2*1)=56.
Good luck on your exam!
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