<PRE><B>Question 75537</B>
In a recent study 90 percent of the homes in the United States were found to have color TVs. In a sample of nine homes, what is the probability that: 
p(have color TV)=0.9
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a. All nine have color TVs?
0.9^9=0.3874
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b. Less than five have color TVs?
binomcdf(9,0.9,4)=0.0008909
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c. More than five have color TVs? 
1-binomcdf(9,0.9,5)=0.99166
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d. At least seven homes have color TVs?
1-binomcdf(9.0,9,6)=0.94703
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Cheers,
Stan H. 
</PRE>