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:
![P(X=x)={n\choose x}p^x(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5Ex%281-p%29%5E%7Bn-x%7D)
#Given n=15, p=0.56, the probability of none will order a non-alcoholic drink:
![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](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5Ex%281-p%29%5E%7Bn-x%7D%5C%5C%5C%5C%5C%5CP%28X%3D0%29%3D%7B15%5Cchoose%200%7D0.56%5E0%281-0.56%29%5E%7B15%7D%5C%5C%5C%5C%5C%5C%3D0.0000045)
![\approx 0.0000](https://tex.z-dn.net/?f=%5Capprox%200.0000)
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:
![P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 4)=1-P(X](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5Ex%281-p%29%5E%7Bn-x%7D%5C%5C%5C%5CP%28X%5Cgeq%20%204%29%3D1-P%28X%3C4%29%5C%5C%5C%5C%5C%5CP%28X%3D0%29%3D1-%5B%7B15%5Cchoose%200%7D0.56%5E0%281-0.56%29%5E%7B15%7D%2B%7B15%5Cchoose%201%7D0.56%5E1%281-0.56%29%5E%7B14%7D%2B%7B15%5Cchoose%202%7D0.56%5E2%281-0.56%29%5E%7B13%7D%2B%7B15%5Cchoose%203%7D0.56%5E3%281-0.56%29%5E%7B12%7D%5C%5C%5C%5C%5C%5C%3D1-%5B0.0000%2B0.0001%2B0.0008%2B0.0042%5D%5C%5C%5C%5C%3D0.9949)
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:
![P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5Ex%281-p%29%5E%7Bn-x%7D%5C%5C%5C%5CP%28X%3C5%29%3DP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%2BP%28X%3D3%29%2BP%28X%3D4%29%5C%5C%5C%5C%5C%5CP%28X%3C5%29%3D%7B15%5Cchoose%200%7D0.56%5E0%281-0.56%29%5E%7B15%7D%2B%7B15%5Cchoose%201%7D0.56%5E1%281-0.56%29%5E%7B14%7D%2B%7B15%5Cchoose%202%7D0.56%5E2%281-0.56%29%5E%7B13%7D%2B%7B15%5Cchoose%203%7D0.56%5E3%281-0.56%29%5E%7B12%7D%5C%5C%5C%5C%5C%5C%3D0.0000%2B0.0001%2B0.0008%2B0.0042%2B0.0161%5C%5C%5C%5C%3D0.0212)
Hence, the probability of less than 5 orders is 0.0212
d. The probability of all orders being non-alcoholic is equivalent to 1 minus no order being non-alcoholic.
-From a above, the probability of zero non-alcoholic order is , P(X=0)=0000045
-Therefore:
![P(All)=1-P(none)\\\\=1-0.0000045\\\\=0.999996](https://tex.z-dn.net/?f=P%28All%29%3D1-P%28none%29%5C%5C%5C%5C%3D1-0.0000045%5C%5C%5C%5C%3D0.999996)
![\approx 1.0000](https://tex.z-dn.net/?f=%5Capprox%201.0000)
Hence, the probability that all orders are non-alcoholic 1.0000