A brick manufacture test the strenght of their bricks by breaking them and deter

Question

A brick manufacture test the strenght of their bricks by breaking them and determing how much pressure was needed to do so. They can not test every brick. In the batch of brikcs on a given day, they test 20 bricks. If more than 2 bricks are found to be defective, they scrap the entire load. a) Suppose that one day's load contacins 5% bad bricks. What is the probability the load will be accepted? b) suppose that on another day 20% of the bricks are bad. What is the probability the load will be rejected?  

Explanation / Answer

I used a calculator rather than a table, but it's to the same effect. Using the binomial: (20 C 1)*(1/20)^1*(19/20)^19-(19/20)^20=0.7356. You were right. I was reading rejected for it, not accepted. 1-(20 C 1)*(1/5)^1*(4/5)^19-(4/5)^20=0.9308247 make sense?

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