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anggar20
29.07.2019 •
Mathematics
Alot of 100 semiconductor chips contains 15 that are defective. three chips are selected at random from the lot without replacement. what is the probability that the third chip is defective?
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Ответ:
d=event that chip selected is NOT defective
Four possible scenarios for the first two selections:
P(DDD)=15/100*14/99*13/98=13/4620
P(DdD)=15/100*85/99*14/98=17/924
P(dDD)=85/100*15/99*14/99=17/924
P(ddD)=85/100*84/99*15/98=17/154
Probability of third selection being defective is the sum of all cases,
P(XXD)=P(DDD)+P(DdD)+P(dDD)+P(ddD)
=3/20
Ответ:
267.9 in^3
Step-by-step explanation: