Question number 11, in how many different ways is it possible to throw a six-sided die three times so that at least one of the throws shows two dots and the sum of the numbers obtained in the first two throws equals the number obtained in the third throw? What we note before enumerating all the possibilities is that the order does matter. So for example, if the first throw is a two and the second throw is a one, then the sum is three and that is a possibility for the third throw. However, reversing these, throwing first a one and then a two would count as a different event. So these are two events. And now we have to make sure that one of the numbers here is always a two and try to partition die face values into a two and a second number. So the first case that we can list above here is one plus one. That equals to two. That's a possibility. And then in the subsequent higher throws, we have to keep a two. So two plus three, for example, is the next possibility. That would be a five. Three plus two is a different event. That would be a five. And then we have two plus four, which would be a six. A four plus two, which would be a six. And let's not forget two plus two, which would give us a four. And those are all the events. If we increase four to five, then we would exceed the number of dots on any face. We would have a sum of seven, which is not possible. So these are all the events, and we can count them. There are eight of them over here. So a total of eight. And we are done counting. So that is the answer, D.

dice throws

six-sided die

probability

combinations

unique sequences