Worked example: counters in bags
This worked example computes the number of ways we can move counters around in two bags to determine a probability of interest.
Problem
Now we have reviewed the key concepts behind Probability in STEP, lets test ourselves with an example. Below is Question 13 from STEP II 2008:
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This question is a particularly useful one for illustrating that often, all we need to do is count.
Let's begin by determining what we need to calculate in part (i): how can the single black counter that begins in bag
So, viewing this as a counting problem we now need to think about the number of ways these two scenarios can occur, out of all possible counter moves. Let's begin with the latter: how many ways can
And how many total possible selections of
Putting these two results together we have:
after some simplifying.
Now we turn out attention to the former scenario. To begin, let's observe that this probability splits up nicely as:
So what's the probability we choose
Finally, what is the probability that we choose
Thus the probability of the first scenario of interest is:
Thus, we can finally compute our original required probability! We have:
Before we move on to part (ii) we're tasked with determining for what value of
and so again we find
So, now on to part (ii). Here we have two black counters: one in each Bag, and need to compute the probability they end up in the same Bag. Proceeding as in part (i), we consider how this can happen. Now we have three scenarios contributing to this probability:
- Scenario 1: The black counter in Bag
was moved to and then both moved back to , - Scenario 2: The black counter in Bag
was moved to and then both remain in , - Scenario 3: The black counter in Bag
is not moved and the black counter from Bag is moved to .
Computing the probability for each of these scenarios is then done by counting exactly like in part (i). You should be able to convince yourself that we have:
Thus, our required probability is:
To finish, we again need to compute the value of maximises this probability. Here, we turn to differentiation:
Since the numerator above is a negative quadratic in k, our maxima of interest will occur at the positive root, i.e. at . But, this isn't an integer; and must be. Thus, our actual to maximise the probability will be one of the integers either side of this root.
And that's the question done! The counting involved was indeed complicated, but as noted earlier, it was only that: counting. With practice you can quickly become more adept at computing such probabilities yourself, and once you are, probability questions can definitely be your friend in STEP.
Thus, our required probability is:
To finish, we again need to compute the value of
Since the numerator above is a negative quadratic in k, our maxima of interest will occur at the positive root, i.e. at
And that's the question done! The counting involved was indeed complicated, but as noted earlier, it was only that: counting. With practice you can quickly become more adept at computing such probabilities yourself, and once you are, probability questions can definitely be your friend in STEP.