Given:
probability of ordering non-alcoholic beverage = 0.48
probability of not ordering non-alcoholic beverage = 1 - 0.48 = 0.52
FInd: the probability that in a sample of 12 customers, at least 5 will order a nonalcoholic beverage.
Solution:
Recall the binomial probability formula.
![P(x)=nCr* p^r* q^(n-r)](https://img.qammunity.org/2023/formulas/mathematics/college/mvp4f8kgl42b8lqb6fb1e714e1u5jrguzi.png)
where
p = probability of success: 0.48
q = probability of failure: 0.52
n = the number of samples: 12
r = number of success (at least 5 which means not 1, 2, 3, or 4.
To determine the probability of having at least 5, let's calculate when r = 0, r = 1, r = 2, r = 3, and r = 4.
Let's start with r = 0 and solve.
![P(0)=_(12)C_0*0.48^0*0.52^(12)](https://img.qammunity.org/2023/formulas/mathematics/college/il0a1y6zi9rfgd9cbv1kl7npfz63gkmogh.png)
![\begin{gathered} P(0)=1*1*0.000390877 \\ P(0)=0.000390877 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rxj1cd5rnn3yr4bb1hfkw4bffxkvozwa5o.png)
At r = 1,
![P(1)=_(12)C_1*0.48^1*0.52^(11)](https://img.qammunity.org/2023/formulas/mathematics/college/ifjzrb6myx9xpxrirgk08mlisshpp6cmmb.png)
![\begin{gathered} P(1)=12*0.48^*0.000751686 \\ P(1)=0.0043297 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qobcyq0w8pawhn5dqgqscwy5eu9sd0q6lm.png)
Now, let's solve for r = 2.
![P(2)=_(12)C_2*.48^2*.52^(10)](https://img.qammunity.org/2023/formulas/mathematics/college/re7awdh2hyafb856lao4kh14ci1k3pe524.png)
![\begin{gathered} P(2)=66*.2304*.00144555 \\ P(2)=0.02198 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ajlnx86f8rf64yii25yb2vg5yx333n455p.png)
Moving on to r = 3.
![P(3)=_(12)C_3*0.48^3*0.52^9](https://img.qammunity.org/2023/formulas/mathematics/college/qtj0q861rpnt9y9v8zcah5u8ep7zwazymv.png)
![\begin{gathered} P(3)=220*0.110592*0.0027799 \\ P(3)=0.067636 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iv4tv3xrerf8fb8jv5ui5r5dy1hehf6m7h.png)
Then, lastly at r = 4.
![P(4)=_(12)C_4*0.48^4*0.52^8](https://img.qammunity.org/2023/formulas/mathematics/college/pb49rho3xv09izrxhw8ce78vugo65mx8ja.png)
![\begin{gathered} P(4)=495*0.05308*0.00534597 \\ P(4)=0.14047 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d8edc9gopfj4bwvgc8sihx1e182cet2179.png)
Let's now add the probability of getting r = 0, r = 1, r = 2, r =3, and r = 4 customers ordering a nonalcoholic beverage.
![P(0)+P(1)+P(2)+P(3)+P(4)](https://img.qammunity.org/2023/formulas/mathematics/college/901xsg6pffwz1ypz37vn1d0j63av70vyzc.png)
![0.0043297+0.02198+0.067636+0.14047=0.2344157](https://img.qammunity.org/2023/formulas/mathematics/college/u2q3cr4g4rcpbu086twip4jwgmhepih4qn.png)
![0.000390877+0.0043297+0.02198+0.067636+0.14047=0.2348](https://img.qammunity.org/2023/formulas/mathematics/college/5gg88gkkwm6co1rcs3wr1qyyzesezb42tn.png)
0.2348 is the probability of at most 4 customers ordering a non-alcoholic beverage.
Since the question is the probability of at least 5 customers ordering a non-alcoholic beverage which is the opposite of the at most 4 customers, then, let's subtract its probability from 1.
![1-0.2348=0.7652](https://img.qammunity.org/2023/formulas/mathematics/college/zp8h6g892kvdperdud6cxh6s7pw84qmtys.png)
Therefore, the probability that in a sample of 12 customers, at least 5 will order a nonalcoholic beverage is approximately 0.7652.