SOLUTION: a)A quality control engineer inspects a random sample of three batteries from a lot of 30 car batteries, which are ready to be shipped. If such a lot contains ten batteries with sl

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Question 1197318: a)A quality control engineer inspects a random sample of three batteries from a lot of 30 car batteries, which are ready to be shipped. If such a lot contains ten batteries with slight defects,
what then is the probability that the inspector's sample contain
i. none of the batteries with slight defects?
i i. at least one of the batteries with slight defects?
B)The probability that a bomber hits its target on any particular run is 80%. If three bombers sent out after the same target, what then is the probability that at least one of the three bombers will hit the target? [Marks -5]
Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi 
Binomial Probability:  n = 30, p = 3/30 = .1
what then is the probability that the inspector's sample contains:
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P( x = 0) = .0424
0r
P+%28x%29=+highlight_green%28nCx%29%28p%5Ex%29%28q%29%5E%28n-x%29+ 
P( x = 0) = 1(.1^0)(.9^30) = .0424
-----------------
n = 3, p = .8
P(x = 0) = .008
P(at least one) = 1 - P(0) == 1 - .008 = .992
Wish You the Best in your Studies.


Answer by ikleyn(52818) About Me  (Show Source):
You can put this solution on YOUR website!
.
a) A quality control engineer inspects a random sample of three batteries from a lot
of 30 car batteries, which are ready to be shipped. If such a lot contains ten batteries
with slight defects, what then is the probability that the inspector's sample contain
i. none of the batteries with slight defects?
ii. at least one of the batteries with slight defects?
b) The probability that a bomber hits its target on any particular run is 80%.
If three bombers sent out after the same target, what then is the probability
that at least one of the three bombers will hit the target? [Marks -5]
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            The solution by @ewatrrr for the first problem  (a)  is  WRONG  for both  (i)  and  (ii)
            not only numerically,  but  CONCEPTUALLY.

            This problem  IS  NOT  on binomial distribution.

            I came to bring you a correct solution. 


(i) The probability in this problem is the ratio of two numbers.

    The denominator is the number of all possible triples of batteries from 
    the whole set of 30 batteries.  The number of such triples is the number of all combinations

        C%5B30%5D%5E3 = %2830%2A29%2A28%29%2F%281%2A2%2A3%29 = 4060.


    The numerator of this fraction is the number of all triples comprising of good batteries.
    It is the number of combinations  C%5B30-10%5D%5E3 = C%5B20%5D%5E3 = %2820%2A19%2A18%29%2F%281%2A2%2A3%29 = 1140.


    Thus the probability  P(i) = C%5B20%5D%5E3%2FC%5B30%5D%5E3 = 1140%2F4060 = 57%2F203 = 0.2808  (rounded).    ANSWER



(ii)  In this problem, the probability is the COMPLEMENT to what we found in (i)

         P(ii) = 1 - P(i) = 1 - 0.2808 = 0.7192   (rounded).    ANSWER

Solved.

Next problem was solved several times by different tutors today, so I do not touch it.


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In the future,  please do not post more than one problem per post.

It is the  REQUIREMENT,  the  RULE  and the  POLICY  of this forum.

Do not make this stupid mistake.