Exercise 103 Page 685

Practice makes perfect

a A petit four is made from 3 of 4 different layers of filling. The pastry chef has 12 fillings to choose from. As the cakes are eaten in a single bite, we do not need to worry about the order of the fillings. For this reason, each selection of fillings is a combination.

The number of combinations _n C_r is the number of ways we can choose r elements from n elements when the selection order is not important.

In our case, n= 12, since we choose the fillings from 12 possible, and r= 3 or r = 4 since each petit four has 3 of 4 fillings. This means the chef made _(12) C_3 petit fours with 3 fillings and _(12) C_4 petit fours with 4 fillings. We can put this expression into a calculator to get its value. _(12)C_3 + _(12)C_4 = 715

b We want to find the probability of getting a petit four that has both raspberry and custard fillings. To do so, let's find the number of ways the cake with these two fillings can be created. As the cakes are eaten in a single bite, we do not need to worry about the order of the fillings. For this reason, each selection of fillings is a combination.

The number of combinations _n C_r is the number of ways we can choose r elements from n elements when the selection order is not important.

We want to focus on these cakes that have raspberry and custard fillings. This means we have 12- 2 = 10 fillings to choose from, and we need 3-2 = 1 or 3-2= 2 fillings to finish the cake. So, we have n = 10 and r = 1 or r= 2. Overall, we have the following number of cakes with raspberry and custard fillings.

_(10) C_1 + _(10) C_2 Let's put this expression into a calculator to get its value. _(10) C_1 + _(10) C_2 = 55 In Part A, we found that the total number of petit fours that the chef made is 715. Let's divide the number of cakes with the two desired fillings by the total number of petit fours to get the desired probability. 55/715 = 0.076923... ≈ 7.69 %