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 cake. Work out the probability that he takes two...
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 cake. Work out the probability that he takes two different cakes.
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.