Since tea and coffee sales are independent the joint distribution of their sold amounts in a minute is the product of their individual distributions.
Coffee sales are Poisson w/mean $\lambda_c=1.5$
Tea sales are Poisson w/mean $\lambda_t=0.5$
So the Joint PMF is given by
$p_{C,T}(c,t)=\left(\dfrac{(1.5)^c}{c!}e^{-1.5}\right)\left(\dfrac{(0.5)^t}{t!}e^{-0.5}\right)$
for (i) simply plug (1,1) in for (c,t)
for (ii) just recast things to be over the course of 3 minutes rather than 1.
Coffee sales are now Poisson with $\lambda_c=3*1.5=4.5$
Similarly Tea sales are now Poisson with $\lambda_t=1.5$
You should be able to finish from here.
hi , i am now having problem with part II
less than 5 cups drink sold= 4 cup + 3 cup + 2 cup +1 cup + 0 cup, ( I let coffee = X , tea = Y) ,
so i have ( 3X , 1Y ) + ( 3Y , 1X ) +( 2X , 1Y) +( 2Y , 1X) + ( 1X , 1Y) +(2X , 0Y) + (2Y , 0X) + (1X) + (1Y) = 0.877 , please refer to the photo below for the working.