1. The problem statement, all variables and given/known data A coffee shop sell tea and coffee. The number of cups of coffee sold in a minute can be assumed to be a random poisson variable with mean = 1.5 . The number of cups of tea sold can be assumed to be an independent random variable with mean = 0.5. Calculate the probablity that in a period of 3minutes , less than 5 cups of drink were sold. 2. Relevant equations 3. The attempt at a solution 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. sorry , since this involve complicated maths formula , so i cant type it one by one and post it here . i attached the working in the photo .