16 Management for a chain of restaurants recorded the number

16. Management for a chain of restaurants recorded the number of appetizers, X, ordered by tables dining. They observed that X had the following probability distribution. Value of X 0 1 2 3 or more Probability 0.60 0.35 0.04 0.01 The probability that a randomly chosen table orders at least one appetizer is 0.35. 0.39. 0.40. None of the above

Solution

given

X                    0             1           2               3 or more

probability      0.60          0.35        0.04         0.04

where X is number of appetizer ordered by dining table

we have to find the probability that a randomly chosen table orders at least one appetizer

means P(X>1)=?

then possible number of appetizer is X=1 or 2 or 3 or more

so

P(X>1) =P(X=1)+P(X=2) +P(X=3 or more)

           =0.35+0.04+0.01=0.40 answer