There are 12 balls in a bag that are numbered 1-12. I will choose one number (say I choose no. 5). Then four balls are picked at random. What is the probability of one of the balls picked being the number I chose?

How will the probability change if I choose two, or three numbers?

I was thinking: there is a 1/12 chance that the first ball i pick could be the number I selected (i.e. 5) and 1/11 chance in the next ball...etc. so if i only chose one number, then my probability would be 1/12 x 1/11 x 1/10 x 1/9 = 1/11880

if i selected two numbers, then the probability would be 2/12 x 2/11 x 2/10 x 2/9 = 2/1485.

I tried to improve my chances by selecting more numbers. However, if I decided to improve my chance another way- by picking out more balls. Then the formula will fail because- I choose one number (ie. 5) and I pick five balls, 1/12 x 1/11 x 1/10 x 1/9 x 1/8 = 1/95040! The probability actually decreases if i try to improve my chances.

I need help with this.

It would be of great help if someone could detail the variations in the probabilities with changes to the number of numbers selected at the beginning. THANKS SO MUCH.

It is often useful to invert problems like these by asking what the probability is of *not* meeting the required condition and subtracting this probability from one.

For the problem at hand, the probability of *not* drawing a ball with the selected number is as follows:
The first ball drawn can be any one of 11 out of 12, P = 11/12
The second ball can be any one of 10 out of 11, P = 10/11
The third ball can be any one of 9 out of 10, P = 9/10
The fourth ball can be any one of 8 out of 9, P = 8/9.

Since all four sub-conditions must be met to *not* draw the correct ball, the overall probability of *not* drawing the ball with the correct number is given by:
P = 11/12×10/11×9/10×8/9 = 8/12 = 2/3.

Therefore, the probability of drawing the correct ball is P = 1–2/3 = 1/3.

You can use a similar approach to calculate the probabilities associated with more numbers (and/or a different number of balls drawn). So, for two numbers:
P = 1–10/12×9/11×8/10×7/9 = 19/33 (<- this is the probability that at least one of the balls drawn has a selected number, and includes the case where both do).

My pleasure. Below is an alternative approach to the problem that shows how it may be better understood from elementary principles. Also, the solutions are shown for the case where *all* of the numbers must match, i.e. in the case of two numbers that both of them appear among the balls picked.

For 1 number, the 4 possible ways of matching it from 4 draws with their associated probabilities are:
(1): P = (1/12)×(11/11)×(10/10)×(9/9) = 1/12
(2): P = (11/12)×(1/11)×(10/10)×(9/9) = 1/12
(3): P = (11/12)×(10/11)×(1/10)×(9/9) = 1/12
(4): P = (11/12)×(10/11)×(9/10)×(1/9) = 1/12
==> P = 1/12+1/12+1/12+1/12 = 4/12 = 1/3

For 2 numbers, the 6 possible ways of matching both from 4 draws with their associated probabilities are:
(1, 2): P = (2/12)×(1/11)×(10/10)×(9/9) = 1/66
(1, 3): P = (2/12)×(10/11)×(1/10)×(9/9) = 1/66
(1, 4): P = (2/12)×(10/11)×(9/10)×(1/9) = 1/66
(2, 3): P = (10/12)×(2/11)×(1/10)×(9/9) = 1/66
(2, 4): P = (10/12)×(2/11)×(9/10)×(1/9) = 1/66
(3, 4): P = (10/12)×(9/11)×(2/10)×(1/9) = 1/66
==> P = 1/66+1/66+1/66+1/66+1/66+1/66 = 6/66 = 1/11

For 3 numbers, the 4 possible ways of matching all 3 from 4 draws with their associated probabilities are:
(1, 2, 3): P = (3/12)×(2/11)×(1/10)×(9/9) = 1/220
(1, 2, 4): P = (3/12)×(2/11)×(9/10)×(1/9) = 1/220
(1, 3, 4): P = (3/12)×(9/11)×(2/10)×(1/9) = 1/220
(2, 3, 4): P = (9/12)×(3/11)×(2/10)×(1/9) = 1/220
==> P = 1/220+1/220+1/220+1/220 = 4/220 = 1/55

For 4 numbers, the only possible way of matching all 4 from 4 draws with its associated probability is:
(1, 2, 3, 4): P = (4/12)×(3/11)×(2/10)×(1/9) = 1/495

Please feel free to ask specific questions if anything above is not clear. The important thing here is to realise some of the abstract principles that are at work in counting problems such as this one.