Answer:
a. 0.0000
b. 0.9949
c. 0.0212
d. 1.0000
Step-by-step explanation:
a. This is a binomial probability distribution problem of the form:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}[/tex]
#Given n=15, p=0.56, the probability of none will order a non-alcoholic drink:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\\\P(X=0)={15\choose 0}0.56^0(1-0.56)^{15}\\\\\\=0.0000045[/tex]
[tex]\approx 0.0000[/tex]
Hence, the probability that none will order a non-alcoholic drink is 0.0000
b. The probability that at least 4 will order a non-alcoholic drink is:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 4)=1-P(X<4)\\\\\\P(X=0)=1-[{15\choose 0}0.56^0(1-0.56)^{15}+{15\choose 1}0.56^1(1-0.56)^{14}+{15\choose 2}0.56^2(1-0.56)^{13}+{15\choose 3}0.56^3(1-0.56)^{12}\\\\\\=1-[0.0000+0.0001+0.0008+0.0042]\\\\=0.9949[/tex]
Hence, the probability of at least 4 non-alcoholic orders is 0.9949
c. The Probability that fewer than 5 orders will be made is calculated as:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X<5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\\\\\\P(X<5)={15\choose 0}0.56^0(1-0.56)^{15}+{15\choose 1}0.56^1(1-0.56)^{14}+{15\choose 2}0.56^2(1-0.56)^{13}+{15\choose 3}0.56^3(1-0.56)^{12}\\\\\\=0.0000+0.0001+0.0008+0.0042+0.0161\\\\=0.0212[/tex]
Hence, the probability of less than 5 orders is 0.0212
d. The probability of all orders being non-alcoholic is equivalent to 1 Â minus no order being non-alcoholic.
-From a above, the probability of zero non-alcoholic order is , P(X=0)=0000045
-Therefore:
[tex]P(All)=1-P(none)\\\\=1-0.0000045\\\\=0.999996[/tex]
[tex]\approx 1.0000[/tex]
Hence, the probability that all orders are non-alcoholic 1.0000
