Solving Probability Problem: 100 Diodes, 5% Broken

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The discussion focuses on calculating the probability of a pile of 100 diodes, with 5% broken, being declared valid after checking half of them. The calculation involves determining the combinations of "ok" and broken diodes in the sample of 50 checked diodes. The initial probability was calculated as p(A) = 0.5, but further analysis revealed a corrected probability of p(A) = 0.181. Additionally, the probability of the pile being declared "not valid" was also discussed, emphasizing that the two probabilities should sum to one. The calculations confirm the likelihood of having 0 or 1 broken diode in the sample for the pile to be considered valid.
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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?
 
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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.
 
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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