A traffic control engineer reports that 75% of the vehicles passing through a checkpoint
Homework Answers
P(X=x)=xC5*(0.25^x)*(0.75^(5-x))
So the probability is
P(X<2)=P(X=0)+P(X=1)
=0C5*(0.25^0)*(0.75^(5-0))+1C5*(0.25^1)*(0.75^(5-1))
=0.6328125
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