Binomial Distribution Probability

lina29
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Homework Statement


A quality control engineer tests the quality of produced computers. Suppose that 5% of computers have defects, and defects occur independently of each other.

A- What is the expected number of defective computers in a shipment of twenty?
B- Find the probability of exactly 3 defective computers in a shipment of twenty.
C- Find the probability that the engineer has to test at least 5 computers in order to find a defective one.


Homework Equations





The Attempt at a Solution



A-1
B- .061 which were correct

For C I got .00257 which was wrong. I just don't understand how I'm supposed to find at least 5 when I don't know the total number since for finding .00257 I used n=20 which I guess I wasn't supposed to since I got it wrong
 
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hi lina29! :smile:
lina29 said:
Suppose that 5% of computers have defects, and defects occur independently of each other.

C- Find the probability that the engineer has to test at least 5 computers in order to find a defective one.

with questions like this, it often helps to write out the opposite, and find the probability of that

so how would you write (in english) the opposite of "the engineer has to test at least 5 computers in order to find a defective one" ? :wink:
 
the engineer has to test at most 4 computers in order to find a defective one
 
oops!

oops! :redface:

yes, that's right, i asked the wrong question :rolleyes:

i should have asked you just to rewrite the question in terms of successes, rather than failures :smile:
 
lina29 said:

Homework Statement


A quality control engineer tests the quality of produced computers. Suppose that 5% of computers have defects, and defects occur independently of each other.

A- What is the expected number of defective computers in a shipment of twenty?
B- Find the probability of exactly 3 defective computers in a shipment of twenty.
C- Find the probability that the engineer has to test at least 5 computers in order to find a defective one.


Homework Equations





The Attempt at a Solution



A-1
B- .061 which were correct

For C I got .00257 which was wrong. I just don't understand how I'm supposed to find at least 5 when I don't know the total number since for finding .00257 I used n=20 which I guess I wasn't supposed to since I got it wrong

Please show your work. Where, exactly, does the number 0.00257 come from? Without this information it is impossible to help you; we cannot tell if you used the correct formula but made an arithmetical error, or what.

RGV
 
using my calculator I did 1-binomcdf(20,.05,4)
 
I don't know how I would write it in terms of successes rather than failures
 
lina29 said:
C- Find the probability that the engineer has to test at least 5 computers in order to find a defective one.

so how many good ones will there be? :wink:
 
15? assuming that C also is also doing a shipment of 20 computers. So would I find exactly 15 successes?
 
  • #10
lina29 said:
C- Find the probability that the engineer has to test at least 5 computers in order to find a defective one.

if he gets to the 5th computer, what happened to the first 4 ?
 
  • #11
they were nondefective
 
  • #12
yes! :smile:

so probability that the engineer has to test at least 5 computers in order to find a defective one = probability that the first 4 are not defective :wink:

which is … ? :smile:
 
  • #13
(.95)^4=.81451
 
  • #14
ok …

so isn't that the answer?​
 
  • #15
lol I was just making sure it was the right answer before I put it in and it is :) Thanks!
 
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