A restaurant provides two special dinner sets. The demand for the first set is 6

Question

A restaurant provides two special dinner sets. The demand for the first set is 6, 7 or 8 units per day and that for the second set is 2 or 3 units per day. Each unit of the first dinner set generates a profit of $2,000 and each unit of the second set generates a profit of $4,000. Let X and Y be the numbers of units of demand for the first and second dinner sets per day respectively. The following probabilities are known (c) P(X = 6) = 0.25 , P[(X+Y=10) | (Y = 2)] = 0.3 , M(X + YS 10) I (X 27)] = 0.8 Find E) and E(X). Hence find the expected total profit generated by these tweo P(Y = 2) = 0.6 , dinner sets per day (10 marks)

Explanation / Answer

Given that P(X=6) =.25, P(Y=2)=.6,

Given that P(X+Y=10|Y=2)= .3, which basically implies P(X=8).P(Y=2)=.3

The above equation can be views as ( for understanding, sum of two dice rolls is 10 (assume that dice is 10 faced and will show number between 1 to 10, X be the first roll and Y be the 2nd roll). Now, if in second roll, 2 appears and total sum is 10, the no. in 1st roll must be 8

But, since X can have only 3 values (8,7,6)P(X=8)+ P(X=7)+ P(X=6)=1 .5+ P(X=7) +.25=1=P(X=7)=.25

Also, given P(X+Y <=10|X>=7)=.8

Let’s split this equation for X=7 and X=8

But, P(Y=3) + P(Y=2)=1 => P(Y=3)=.4

Hence, E(X)= Summation(x p(x))= 6 P(6) + 7P(7) +8 P(8)è 6 * .25 + 7 * .25 + 8 * .5= 1.5 + 1.75 +4 = 7.25

Similarly, E(Y)= 2P(2) + 3P(3) = 2* .6 + 3* .4 =1.2 + 1.2 =2.4

Hence, profit generated = 2000 * E(X) = 2000* 7.25 =14500

Similarly, from 2nd profit =4000* E(Y)= 4000* 2.4 =9600

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