A lottery operator sources its balls from two different manufacturers, A and B.
Manufacturer
A supplies 85% of all balls and has on average 0.05% rejects,
Manufacturer B has 0.2% rejects.
What is the probability that a randomly selected defective ball came from manufacturer B?
Here we can apply Bayes’ theorem directly:
P(B|defective) = [0.002*0.15] / [0.0005*0.85 + 0.002*0.15] ≈ 0.4138 or 41.38%
Also interesting is the probability that a randomly selected ball is defective at all:
P(defective) = 0.002*0.15 + 0.0005*0.85 = 0.000725 or 0.0725%
Probability in Lotto isn’t that complicated
Calculating probabilities for drawing Lotto numbers is not hard; a little imagination is enough. Mentally simulate the draw: imagine drawing just one ball out of 49, then putting it back and starting again. Before each draw, you guess which ball will be next, draw it, then return it to the drum.
Thus the probability is easy to determine. All balls have the same chance of being drawn. The probability that our one-ball guess is correct is exactly 1:49. Colloquially we say “1 in 49,” mathematically it’s 1 divided by 49. The value is about 0.0204, i.e., roughly 2.04%.
So it’s simple to compute the probability of the first ball: always about 2.04%. Previous draws don’t matter. Even if in our thought experiment the same ball were drawn 17 times in a row, its chance on the next draw would again be about 2.04%. The ball has no memory — every draw is independent. (Note: the probability of the compound event “the ball 5 is drawn 100 times in a row” is extremely small; that’s a different probability than a single draw. We now explain this.)
Example 2:
Again imagine drawing a ball. The probability of guessing it right is 1:49. We return the ball and guess again for the second draw — what is the probability of guessing correctly twice in a row?
For the first draw there are 49 possibilities. In our game, for the second draw there are again 49 possibilities. You can compute the probability by counting combinations. If the first draw is, say, ball 4, there are still 49 options for the second draw. Each first draw can be paired with 49 second draws. Thus there are 49 × 49 = 2,401 possible ordered pairs for the two draws in our fictitious setting.
The probability of guessing both drawn numbers correctly is therefore exactly 1:2,401, i.e., about 0.00042 or 0.042%. Note that the probability for the first draw didn’t change — but the probability of the combined event is much smaller.
If we extend our 1-out-of-49 game, with each additional draw the number of ordered combinations multiplies by 49. With six draws there are 49 * 49 * 49 * 49 * 49 * 49 = 13,841,287,201 possibilities. The probability of predicting all numbers in the correct order is therefore about 1 in 14 billion, i.e., 0.0000000072%.
In mathematics we write the probability space size for the ordered six-draw experiment as 49^6 (“49 to the 6th power”).
For further reading, see the rules and a comparison of Lotto versus Eurojackpot probabilities.
