There are 10 cakes: 1 fruit slice, 6 doughnuts and 3 iced buns. Alex takes a cake at random and eats it. He then takes at random a second...

Hello!

Let's solve this by definition of probability, counting all possible events and the desirable ones.

If we remember what cake was eaten first, then there are `10*9 = 90` possible events (the first cake may be any of `10,` the second is any of `9` remaining).

The desirable cases (when the first and the second cakes are different) may be divided into three disjoint subsets: 1) when the first cake is a fruit slice, 2) when the first is a doughnut and 3) when it is an iced bun. We have to count the number of events in these groups and add them.

In the case 1), there is `1` possibility to choose a fruit slice and `9` possibilities to choose a different second cake. Remember `1*9 = 9.`

In the case 2), there are `3` possibilities to choose a doughnut and only `7` to choose a different second cake (`2` remaining doughnuts are not suitable). Remember `3*7 = 21.`

In the case 3) there are `6*(9-5) = 24` possibilities.

The sum is `9+21+24 = 54.`

So the probability in question is `54/90 = 6/10 = 3/5 = 0.6.`

Choose whatever form you prefer.