In a lottery, 6 numbered balls are drawn from a pool of 59. Calculate the probability of scoring a jackpot. There used to be 49 balls in the pool. Calculate by how much the addition of 10 balls has decreased the probability of scoring a jackpot

The question has to do with probability, and the branch of mathematics which models such situations is called probability theory.

A jackpot is, of course, when we correctly predict all the numbers on the balls which will be drawn.

First, forget everything you have learned in your IB Maths, and let’s try solving this problem by intuition.

1^st draw

Imagine a sack with 59 numbered balls in it. You predict the values of 6 of them, close your eyes and take one random ball out of the sack. What is the probability that the number on it is among the 6 you have predicted?

It is of course 6/59.

2^nd draw

You took one ball from the sack, so now there are 58 of them left. Also, it was one of those you predicted you will pull out, so there are only 5 of those left. The probability of a predicted ball being drawn time is 5/58, but the overall probability of arriving at this point is (6/59) * (5/58) (draw a probability tree if you find this difficult to understand).

3^rd draw

Similarly, now there are 57 balls left in the sack, but only 4 of those you have predicted will be drawn. The overall probability of drawing a predicted ball is now 4/57, and the overall probability is (6/59) * (5/58) * (4/57).

Subsequent draws

If we repeat the same process three more times, we will arrive at an overall probability of 6/59) * (5/58) * (4/57) * (3/56) * (2/55) * (1/54), or 1/45057474. If we were to believe this BBC article (http://www.bbc.co.uk/news/blogs-magazine-monitor-26583325), this is 150 times less likely than the probability of being killed by a lightning strike.

You must of course be wondering whether there exists a quicker way to obtain this answer. Notice that:

The balls are drawn in a completely random way

Each draw is completely independent of others

There are only two possible outcomes for a draw – a success, when we predict the number on the ball, and a failure, when we don’t

If you turn to your IB Maths notes, you will discover that our draws fulfil every criterion of a Bernoulli experiment, and that the probability of success in n subsequent Bernoulli experiments is governed by the Binomial distribution.

An analysis of this distribution tells us that the number of ways of arranging n elements out of r is given by n!/(r!(n-r)!), where n! means n factorial, a multiplication of all integers from 1 to n.

Your calculator has a capability to calculate this type of distribution, usually denoted as nCr(r,n). Indeed, if I type nCr(59,6) into my calculator it tells me there are 45057474 ways to arrange 6 items out of 59, and we only predicted one of them so the probability that it is will be drawn is 1/45057474.

Also, now we can easily answer the second part of the question. Now we are drawing 6 balls out of 49, so there are nCr(49,6) = 13983816 ways to arrange them.

So by increasing the number of balls to 59 we increased the number of possible arrangements by 45057474/13983816 or 3.2 times.