SOLUTION: Twenty tires are tested to see if they last as long as they claim they do. Three tire fail the test. Two tires are selected at random without replacement for a detained inspection.

Algebra ->  Probability-and-statistics -> SOLUTION: Twenty tires are tested to see if they last as long as they claim they do. Three tire fail the test. Two tires are selected at random without replacement for a detained inspection.      Log On


   

Question 349882: Twenty tires are tested to see if they last as long as they claim they do. Three tire fail the test. Two tires are selected at random without replacement for a detained inspection.
A. Find the probability that both tires failed the test.
B. Find the probability that both tires passed the test.
C. Find the probability that at least one of the tire passed the test.
Answer by sudhanshu_kmr(1152) About Me  (Show Source):
You can put this solution on YOUR website!
out of 20 tire 3 are defective.
A. no. of ways to select 2 tire from 20 = 20C2
no. of ways to select 2 tire from 3 defective tire = 3C2
required probability = 3C2 / 20C2
= 3/ 190

B.
no. of ways to select 2 tire from 17 non-defective tire = 17C2

required probability = 17C2 /20C2
= 8 * 17 / 190
= 68/ 95

C.
no. of ways to select 1 from 3 defective and
1 tire from 17 non defective= 3C1* 17C1 = 3 * 17 = 51

probability when one tire passed the test = 51/20C2 = 51 /190
now, to find the required probability when at least one tire passed the test,
we will add the probability of both

a. when both tire passed the test
b. when only tire passed the test

required probability = 68/95 + 51/190
= 187 / 190


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