Probability problem

  • Thread starter vabamyyr
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  • #1
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there is a pile of diodes. Half of them are checked. The pile is declared valid, if there are no more than 2% of broken diodes.

Now i have to find the probability when having a pile of 100 diodes and 5% of them are broken, would be declared valid.

How i did this task. there are 95 "ok" diodes and 5 broken diodes. The probability of this set of diodes declared valid is p(A)= (95C48*5C2 + 95C49*5C1 + 95C50*5C0)/100C50 which equals 0.5. The answer after the problem is said to be p(A)= 0,181
Am I thinking wrong?
 

Answers and Replies

  • #2
EnumaElish
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Perhaps as a check for your answer calculate the prob. of being declared "not valid." The two probs. should add to one.
 
  • #3
Pr(5% broken is declared ok/1/2 are checked)=Pr(you get either 0 broken or 1 broken in the pile of 50)

=Pr(you get 0 or 1 faulty in 50/5 are faulty in 100)=50C5+50.50C4/{2*50C5+2*50C4*50+2*50C3*50C2}=0.1810892429
 

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