If there are 15 customers, and X represents the number who choose diet, and Y represents the number who choose non-diet, then X+Y=15. So we only need one variable to describe the number of customers who choose a drink of a particular type.
Since there are only 10 cans of each type available and X+Y = 15, the allowable combinations of (X,Y) are
(5,10),(6,9),(7,8),(8,7),(9,6),(10,5)The random variable X follows binomial distribution with p=0.6 and n=15: X∼B(15,0.6)
b(x;n,p)=(xn)pxqn−x,x=0,1,2,...,n; q=1−p The probability that each of to 15 customers get the drink they want is
P(5≤X≤10)=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)
P(X=5)=(515)0.65(1−0.6)15−5≈0.02448564P(X=6)=(615)0.66(1−0.6)15−6≈0.06121411P(X=7)=(715)0.67(1−0.6)15−7≈0.11805577P(X=8)=(815)0.68(1−0.6)15−8≈0.17708366P(X=9)=(915)0.69(1−0.6)15−9≈0.20659761P(X=10)=(1015)0.610(1−0.6)15−10≈0.18593784
P(5≤X≤10)≈0.02448564+0.06121411+0.11805577++0.17708366+0.20659761+0.18593784≈≈0.773375 (to 6 decimal places) The probability that each of the 15 is able to purchase the type of drink desired is 0.773375 (about 77.3%).
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