Question 1171774: Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine if the events are unusual. If convenient, use the appropriate probability table or technology to find the probabilities. 47 percent of adults say that they have cheated on a test or exam before. You randomly select 6 adults. Find the probability that the number of adults who say that they have cheated on a test or exam before is (a) exactly 4 , (b) more than 2, and (c) at most 5.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a binomial distribution because there is a fixed probability, fixed finite sample, and the probability doesn't change for each selection. Poisson would have unlimited (potentially) events, and geometric would be looking for the probability that the first event of interest would occur on the third or some other fixed trial.
a. 6C4*0.47^4*0.53^2=0.2056
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b. Look at 0,1,2 and take the complement
for 0 it is 0.53^6=0.0222
for 1 it is 6*.47*.53^5=0.1179
for 2 it is 0.2615
They sum to 0.4016
The answer is the complement or 0.5984
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c. For at most 5, want to get 6 and subtract that from 1.
prob (6)=0.47^6=0.01078
the answer is 0.9892
For the answers to a, b, c, none of these events is unusual.
For 0 or for 6, it would be uncommon, since the probability is almost 1/2 and that would be 1/64 probability approximately.
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