Thus far, we have mostly talked about problems where all of the outcomes are equally likely. So, for example, our coins are equally likely to land heads as they are to land tails, and our spinners are equally likely to land on any of the outcomes, and our die are equally likely to land on any of 1 through 6. But not all GRE problems are like this. Consider the following:
Suppose you have a weighted coin which is twice as likely to land tails as it is to land heads. You are about to flip this coin. What is the likelihood of it landing heads?
In these problems, we need to do a little bit of algebra to find the answer. Feel free to think about it, and if you want to see a solution, click below:
We know that P(Heads) has some value. Let's call that value . So we write: . And, by what the question states, the coin is twice as likely to land on tails as it is to land on heads. So,
Now, we know that P(Tails) + P(Heads) = 1, since it is guaranteed that the coin lands on heads or tails and the two events are mutually exclusive (see our post on P(A or B) if this reasoning is unfamiliar). Thus,
By division, we conclude that
So in questions where the outcomes are unequal, we can use variables as placeholders for the true probabilities and then use the probability rules to derive the actual probabilities of the different outcomes. Here are some practice problems that are basically variations on this theme:
Suppose you have a weighted six-sided die that is four times as likely to land on the number 4 as it is to land on any of the other numbers. What is the probability that it lands on 6?
Let denote the event of the dice landing on 4 (and similarly for all the other numbers). Let Then, We know that since the die has to land on one of the numbers and all of them are mutually exclusive. Thus, we know
Thus, so The probability that the die lands on 6, then, is just
Suppose you have a company raffle. Managers (of which there are 30) get 2 tickets, ordinary employees (of which there are 110) get 1 ticket, and executives (of which there are 5) get 20 tickets. What is the probability that the winner is an executive?
At first, this also looks like a problem where different outcomes have different likelihoods. The chances of a particular executive winning are not equal to the chances of a particular manager winning (since executives get more tickets than managers). But we can think about the problem in a slightly different way and thereby turn it into a problem where the outcomes are equally likely. Thus, we can consider the outcomes for each tickets. There are 100 executive-tickets, 110 ordinary employee tickets, and 60 manager-tickets. Each ticket is equally likely to win, so there is a chance that an executive-ticket wins. In other words, there is a chance that an executive wins.
Suppose you have an 8-sided spinner numbered 1 through 8. Now, your spinner is set up so that you are twice as likely to get 2 as you are to get 1, 3 times as likely to get 3 as your are to get 1, and so on for all the other numbers up until 8. What is the probability that you spin a 1?
Let Then, and and so on for the rest. Thus,
Now, adding Thus, and so