Answer:
b. 0.864
Step-by-step explanation:
Let's start defining the random variables.
Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''
Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''
If X is a binomial random variable, the probability function for X is :
![P(X=x)=(nCx)p^{x}(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%28nCx%29p%5E%7Bx%7D%281-p%29%5E%7Bn-x%7D)
Where P(X=x) is the probability of the random variable X to assume the value x
nCx is the combinatorial number define as :
![nCx=\frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=nCx%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
n is the number of independent Bernoulli experiments taking place
And p is the success probability.
In counter I :
Y1 ~ Bi (n,p)
Y1 ~ Bi(2,0.2)
![P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}](https://tex.z-dn.net/?f=P%28Y1%3Dy1%29%3D%282Cy1%29%280.2%29%5E%7By1%7D%280.8%29%5E%7B2-y1%7D)
With y1 ∈ {0,1,2}
And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}
In counter II :
Y2 ~ Bi (n,p)
Y2 ~ Bi (1,0.3)
With y2 ∈ {0,1}
And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}
(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :
![P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}](https://tex.z-dn.net/?f=P%28Y2%3Dy2%29%3D%280.3%29%5E%7By2%7D%280.7%29%5E%7B1-y2%7D)
Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.
![P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)](https://tex.z-dn.net/?f=P%28Y1%3Dy1%2CY2%3Dy2%29%3DP%28Y1%3Dy1%29.P%28Y2%3Dy2%29)
![P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}](https://tex.z-dn.net/?f=P%28Y1%3Dy1%2CY2%3Dy2%29%3D%282Cy1%29%280.2%29%5E%7By1%7D%280.8%29%5E%7B2-y1%7D%280.3%29%5E%7By2%7D%280.7%29%5E%7B1-y2%7D)
With y1 ∈ {0,1,2} and y2 ∈ {0,1}
P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}
b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :
![Y1 + Y2 \leq 1](https://tex.z-dn.net/?f=Y1%20%2B%20Y2%20%5Cleq%201)
when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1
To calculate
we must sume all the probabilities that satisfy the equation :
![P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)](https://tex.z-dn.net/?f=P%28Y1%2BY2%5Cleq%201%29%3DP%28Y1%3D0%2CY2%3D0%29%2BP%28Y1%3D1%2CY2%3D0%29%2BP%28Y1%3D0%2CY2%3D1%29)
![P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448](https://tex.z-dn.net/?f=P%28Y1%3D0%2CY2%3D0%29%3D%282C0%29%280.2%29%5E%7B0%7D%280.8%29%5E%7B2-0%7D%280.3%29%5E%7B0%7D%280.7%29%5E%7B1-0%7D%3D%280.8%29%5E%7B2%7D%280.7%29%3D0.448)
![P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224](https://tex.z-dn.net/?f=P%28Y1%3D1%2CY2%3D0%29%3D%282C1%29%280.2%29%5E%7B1%7D%280.8%29%5E%7B2-1%7D%280.3%29%5E%7B0%7D%280.7%29%5E%7B1-0%7D%3D2%280.2%29%280.8%29%280.7%29%3D0.224)
![P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192](https://tex.z-dn.net/?f=P%28Y1%3D0%2CY2%3D1%29%3D%282C0%29%280.2%29%5E%7B0%7D%280.8%29%5E%7B2-0%7D%280.3%29%5E%7B1%7D%280.7%29%5E%7B1-1%7D%3D%280.8%29%5E%7B2%7D%280.3%29%3D0.192)
![P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864](https://tex.z-dn.net/?f=P%28Y1%2BY2%5Cleq%201%29%3D0.448%2B0.224%2B0.192%3D0.864%5C%5CP%28Y1%2BY2%5Cleq%201%29%3D0.864)