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Problem 10

There are $16$ doughnuts on a table. Some of them have a chocolate filling, and some of them have a raspberry filling, and it is impossible to tell which one is which without eating them!

We do know, however, that there are more raspberry doughnuts than chocolate doughnuts.

We also know that if two doughnuts are eaten at random, the probability they have the same filling is $\frac{19}{40}$ .

If the number of chocolate doughnuts is $x$ , and the number of raspberry doughnuts is $y$ , find the value of

x^{3} + y^{2}

Hint

You can use a tree diagram.

Hint

Let the number of chocolate doughnuts be $x$ , then the number of raspberry doughnuts must be $16 - x .$

Solution

Since any removed doughnut must be eaten, we are sampling without replacement.

The probability of selecting chocolate and then chocolate is

\frac{x}{16} \times \frac{x - 1}{15}

Noting that $y = 16 - x$ , the probablity of selecting raspberry and then raspberry is

\frac{y}{16} \times \frac{y - 1}{15} = \frac{16 - x}{16} \times \frac{15 - x}{15}

So the probablity of selecting two doughnuts with the same filling is the sum of these:

\begin{aligned} \frac{x}{16} \times \frac{x - 1}{15} + \frac{16 - x}{16} \times \frac{15 - x}{15} & = \frac{19}{40} \\ \frac{x (x - 1) + (x - 16) (x - 15)}{16 \cdot 15} & = \frac{19}{40} \\ \frac{2 x^{2} - 32 x + 16 \cdot 15}{16 \cdot 15} & = \frac{19}{40} \\ \frac{x^{2} - 16 x + 8 \cdot 15}{8 \cdot 15} & = \frac{19}{40} \\ \frac{x^{2} - 16 x + 8 \cdot 15}{15} & = \frac{19}{5} \\ x^{2} - 16 x + 8 \cdot 15 & = 19 \cdot 3 \\ x^{2} - 16 x + 63 & = 0 \\ (x - 7) (x - 9) & = 0 \\ x & \in {7, 9} \end{aligned}

There are more raspberry than chocolate, so $x = 7$ and $y = 9$ .

Problem 10 ​

Problem 10