Question 1059032: 3. A study estimates that the rate of preventable adverse drug events (ADEs) in hospital is 35.2%. Preventable ADEs typically result from inappropriate care or medication errors, such as errors in commission and omission. Suppose 10 hospital patients experiencing an ADE are chosen at random. Let p=.35 and calculate the probability that:
a. Exactly 7 of those drug events are preventable;
b. None of those drug events were preventable
c. Between 3 and 6 inclusive were preventable.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It is 10C7(0.35)^7*0.65^3=0.0212. This makes sense, since the expected value is 3.5 and the SD is about 1.5 (first is n*p, the second is square root of (np(1-p))
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.65^10=0.0135
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For 3, it is 120(0.35)^3)(0.65)^7=0.2522
For 4, it is 210*(0.35^4)(0.65^6)=0.2377
For 5, it is 252*(0.35)^5*(0.65^5)=0.1536
For 6, it is 210*(0.35^6)(0.65^4)=0.0689
Sum is 0.7124
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