A sample of 4 balls is to be selected at random from an urn containing 18 balls numbered 1 to 18. Five balls are green, 6 balls are white, and 7 balls are black.
(a) How many different samples can be selected?
(b) How many samples can be selected that contain at least 1 white ball?
List the cases in a systematic way:
4 alike
GGGG
WWWW
BBBB
3 alike, 1 different
GGGW
GGGB
WWWB
WWWG
2 alike, 2 alike
GGWW
GGBB
WWBB
2 alike, 2 different
GGWB
WWGB
BBGW
I think I got them all, check them then count them up
To answer these questions, we will use the concept of combinations. The number of combinations can be calculated using the formula:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of items, r is the number of items chosen, and ! denotes factorial.
(a) To find the number of different samples that can be selected, we need to calculate the number of combinations of choosing 4 balls out of 18.
C(18, 4) = 18!/(4!(18-4)!) = 18!/(4!14!)
Simplifying this expression:
18! = 18 x 17 x 16 x 15 x 14!
4! = 4 x 3 x 2 x 1
18! / 4! = (18 x 17 x 16 x 15 x 14!) / (4 x 3 x 2 x 1)
= 3060
Therefore, there are 3060 different samples that can be selected.
(b) To find the number of samples that contain at least 1 white ball, we need to consider two cases: samples with exactly 1 white ball and samples with more than 1 white ball.
Case 1: Samples with exactly 1 white ball
The number of combinations for choosing 1 white ball out of 6, and choosing 3 balls from the remaining 17 (without replacement) is:
C(6, 1) x C(17, 3) = 6!/(1!(6-1)!) x 17!/(3!(17-3)!) = 6 x (17 x 16 x 15)/(3 x 2 x 1) = 680
Case 2: Samples with more than 1 white ball
The number of combinations for choosing 2 white balls out of 6, and choosing 2 balls from the remaining 17 (without replacement) is:
C(6, 2) x C(17, 2) = 6!/(2!(6-2)!) x 17!/(2!(17-2)!) = 15 x (17 x 16)/(2 x 1) = 540
Therefore, the total number of samples that contain at least 1 white ball is:
Case 1 + Case 2 = 680 + 540 = 1220
So, there are 1220 samples that can be selected containing at least 1 white ball.