Winners of a quiz are given the chance to earn tokens, which can be redeemed for prizes. Each token has a value written on the front. Shirley wins the quiz and is told that she must randomly pick one token from each of the two bags shown below. The probability of pulling a token from each bag is fair.
In bag one, we’ve got five tokens: one is labelled one, two are labelled four, one is labelled seven, and one is labelled 10. In bag two, we’ve got four tokens: one is labelled zero, one is labelled three, one is labelled six, and one is labelled 15. And in part a, we have to complete the possibility space diagram for the total value of the tokens that Shirley pulls from the bags.
Now, when we say the probability of pulling a token from each bag is fair, it just means that within any particular bag each token is equally likely to be pulled out. So we’ve also been given the start of a possibility space diagram. We can see across the top, we’ve got the contents of bag one: the tokens one, four, four, seven, and 10. And on the left-hand side, we’ve got the contents of bag two: a token with zero, another with three, another with six, another with 15.
So by combining these together on that grid, we can see what the combined values of the tokens are. If we get a one in bag one and a zero in bag two, the combined total of those two — zero plus one — is one. So to work out the value in any of the remaining squares, we just add together the token value from bag one to the token value from bag two.
If we get a four from bag one and a zero from bag two, four plus zero is four. And that happens in either of these cases. If we have a seven from bag one and a zero from bag two, that’s worth seven and a 10 from bag one and a zero from bag two, 10 plus zero is 10.
Now, combining each of those, we’re getting a three in bag two. Four plus three is seven, seven plus three is 10, and 10 plus three is 13. And we can fill out the other two rows again by adding together the values of the tokens from the two bags.
So our possibility space diagram tells us that there are 20 possible combinations of tokens from bag one and bag two and they add together in these 20 different ways. Let’s make a little bit of space now so we can carry on and do parts b and c of the question.
In part b, we have to find the probability that Shirley pulls two tokens that have a combined value of exactly 10. And this is kind of relative frequency question. The probability that the two tokens sum to 10 is equal to the number of ways that we’ve got of having two tokens that sum to 10 over the number of possible outcomes altogether.
And we saw from our possibility space diagram that there are 20 possible outcomes. So we know the denominator. Now, we’re just gonna work out how many ways are there of picking two tokens which sum to 10. Well, there are four ways: we could get a 10 in bag one and a zero in bag two, we could get a seven in bag one and a three in bag two, or there are two ways of getting a four in bag one and they can combine with a six in bag two.
So that makes our answer four twentieths. Now, we can simplify that to a fifth, if we want to. But with probability questions, we don’t actually have to simplify the answer.
And finally in part c, we have to find the probability that Shirley pulls two tokens which have a combined value of 13 or less. And the method is much the same as in part b. The probability that the two tokens add up to less than or equal to 13 is equal to the number of ways that we’ve got of picking two tokens that sum to less than or equal to 13 divided by the total number of possible outcomes.
Well, we’ve got the same two bags with the same tokens in. So there are still 20 possible outcomes. But we’ve now got to count up how many ways are there of summing to less than or equal to 13. Well, in fact, there are only six ways of getting more than 13. So the other 14 ways all give us a total of 13 or less. So the answer is fourteen twentieths. And again, we could simplify that to seven tenths, if we wanted to. But you don’t have to do that in probability questions.
So for parts b and c, because Shirley was randomly picking and all of those outcomes were equally likely, it just turned into a counting exercise. There were 20 possible outcomes and we had to count how many of the outcomes match the description in each case. So that was four out of 20 in the first case and 14 out of 20 in the second case.