14 Probability

Probability is not a topic in the International Mathematical Olympiad, nor major international competitions generally. In the AMC we did try to set however an interesting probability problem each year, mainly because we found probability the most interesting branch of statistics, and also because, like combinatorics, which we embraced, we felt we could set questions which could be solved without formal knowledge of the theory (even though some of the examples below may look difficult). We even felt we could work round simple Bayesian problems. Statistics did become part of the Australian syllabus from about the time the AMC started, but its syllabus focus would be on aspects such as data analysis. We did frequently set questions on aspects of mean.

I will start with a standard accessible problem from an early AMC.

Example 14.1

Mabel throws two dice (each numbered from 1 to 6), a red one and a white one. What is the probability that the red one beats the white one in score?

Solution 14.1

The probability that the scores are equal is ¹/₆. Hence the probability that the scores differ is ⁵/₆, and so the chance of red winning (by symmetry) is ¹/₂ x ⁵/₆ = ⁵/₁₂.

Wording of probability questions

Probability questions need very much care in wording. They usually describe a real world situation which needs to be very clearly worded to avoid ambiguity. My colleague Roger Curnow was very good at pointing out to me hazards where they existed. An early problem we had set involving the drawing of playing cards had got past Roger and he was critical after he saw it had been set. Roger's point was subtle, and as it turned out no one was confused by the point. However Roger was correct and I checked with him later the following problem which was mathematically similar. On the surface this problem appears to need technical background, including maybe the Bayesian formula, but it can be seen that it can be solved with good street-wise mathematics.

Example 14.2

An honest tennis promoter decides to stage an exhibition match between two players, whose names are to be drawn at random from a pool of N players, n of whom are Australians. He draws the two names and before announcing them is asked by a reporter `Will the match involve at least one Australian?' After looking at the drawn names he says `Yes'. What is the chance of both players being Australian?

Solution 14.2

The chance that both players are Australian

for example, if there are 3 Australians in a pool of 8 players altogether, this is ^{(3 - 1)}/_{(16 - 3 - 1)}=¹/₆.

Comment

This problem was more difficult because it required the formula for all n and N. If these variables had been numbers no bigger that 10 it would have been easier as the cases could all be found manually.

Example 14.3

Two forgetful friends agree to meet in a coffee shop one afternoon but each has forgotten the agreed time. Each remembers that the time was somewhere between 2 pm and 5 pm. Each decides to go to the coffee shop at a random time between 2 pm and 5 pm, wait half an hour, and leave if the other doesn't arrive. What is the probability that they meet?

Solution 14.3 and Comments

This is a very difficult problem to solve, even for an experienced solver, if trying by any sort of symbolic method. In grappling with it one soon understands the situation much better if looking for a graphical-based solution.

We can do this as follows: Let one friend's arrival time be x and the other be y. They can both take random values, say, between 0 and 3. They will meet if |x - y| ≤ ¹/₂. This region is shown below.

Region of |x - y| ≤ ¹/₂.

The probability of their meeting is then

i.e. ¹¹/₃₆.

Comment

I now pose a different problem which I composed for the AMC in its early days.

Example 14.4

Stephanie is in a shop and wishes to buy a 10c chocolate. In her purse she has 12 coins, being 3 each of the 1c, 2c, 5c and 10c denominations. She randomly chooses 3 coins from her purse. What is the probability that the value of these coins is at least enough to pay for the chocolate?

Solution 14.4 Alternative 1

There are

coin combinations altogether. It is easier to count those which are unfavourable, which we do using the following table:

Since the number of unfavourable combinations is

1 + 9 + 9 + 9 + 27 + 1 + 9 = 65

the probability that Stephanie has at least 10 cents is thus

the answer required.

Solution 14.4 Alternative 2

The problem can be considered as an example of selecting at random and without replacement n objects from a total of N objects, N₁ of type I and N₂ of type II (N₁ + N₂ = N; n ≤ N). From combinatoric theory (or the hypergeometric statistical distribution), the probability that the selection results in n₁ items of type I and n₂ items of type II is

where n₁ ≥ 0, n₂ ≥ 0, n₁ + n₂ = n, n₁ ≤ N₁ and n₂ ≤ N₂.

Now Stephanie will not be able to make the 10c purchase if she selects zero or one 5c pieces together with (respectively) three or two 1c or 2c pieces. Hence the probability of Stephanie drawing at least 10c is

where we have divided by the total number of ways of choosing three coins from the 12.