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 :
[tex]P(X=x)=(nCx)p^{x}(1-p)^{n-x}[/tex]
Where P(X=x) is the probability of the random variable X to assume the value x
nCx is the combinatorial number define as :
[tex]nCx=\frac{n!}{x!(n-x)!}[/tex]
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)
[tex]P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}[/tex]
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)
[tex]P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}[/tex]
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 :
[tex]P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}[/tex]
Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.
[tex]P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)[/tex]
[tex]P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}[/tex]
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 :
[tex]Y1 + Y2 \leq 1[/tex]
[tex]Y1 + Y2\leq 1[/tex] when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1
To calculate [tex]P(Y1+Y2\leq 1)[/tex] we must sume all the probabilities that satisfy the equation :
[tex]P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)[/tex]
[tex]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[/tex]
[tex]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[/tex]
[tex]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[/tex]
[tex]P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864[/tex]
