A box contains 20 balls, of which 5 are red, 5 are blue, 5 are white and 5 are green. Suppose a sample of 5 balls are chosen without replacement.
a) find the probability that the sample contains balls of all the same colour
(answer is 4/ 20C5, but why isnt it 5/20C5??)
b) find the probability that the sample contains three balls of one colour and two balls of another colour.
The teacher wrote
5C3 x 5C2 x 4P2
---------------
20C5
Could you please explain why this is?
Thanks!!
a) the ans. is 4/20C5 because there are 4 types of balls and 5 is the magnitude of a specific colour.
b)5C3 gives value of 3 balls of 1 colour;
5C2= value of 2 balls of another colour;
4P2= probality of 2 colours from 4 colours ;
20C5=combination of total balls of different colour.
therefore:-5C3*5C2*4P2
-------------
20C5
Dinga chika
could be all red, all blue , all white or all green
prob(all red) = (5/20)(4/19)(3/18)(2/17)(1/16) = /15504
the same calculation for the others
so the prob(all same colour) = 4(1/15504) = 1/3876
which is the same as 4/C(20,5)
Why isn't it 5/C(20,5) ??
Why would it? There are only 4 colours
3 colours of one kind, 2 of the others :
RRRBB RRRWW RRRGG
BBBRR ...
..
WWWRR ................
There would be 12 of those
let's look at one such case
RRRBB
prob = C(5,3) x C(5,2)/C(20,5) = 10/15504
but there are 12 cases with the same colour triple, and another colour double
so prob = 12 x C(5,3) x C(5,2)/C(20,5)
= 1200/15504
= 25/323
which is the same as your teacher's answer
a) To find the probability that the sample contains balls of all the same color, we need to consider the number of ways we can choose 5 balls of the same color and divide it by the total number of ways we can choose any 5 balls from the box.
There are 4 different colors and to have a sample with balls of all the same color, we can choose any one of these colors. Let's say we choose red.
The number of ways to choose 5 red balls from the box without replacement can be calculated using the combination formula:
5C5 = 1 way to choose all 5 red balls
Now, we need to divide this by the total number of ways we can choose any 5 balls from the box, which can be calculated as:
20C5 = (20! / (5! * (20 - 5)!)) = 15504
So, the probability of choosing 5 balls of the same color is:
1/15504 = 1/20C5
The answer is not 5/20C5 because choosing all red balls is just one possibility out of the 4 colors. We shouldn't count each color separately in the numerator, as that would be overcounting the possibilities.
b) To find the probability that the sample contains three balls of one color and two balls of another color, we need to consider the number of ways we can choose 3 balls of one color, 2 balls of another color, and multiply it by the number of ways to arrange those 5 chosen balls.
Let's say we choose 3 red balls and 2 blue balls.
The number of ways to choose 3 red balls without replacement can be calculated using the combination formula:
5C3 = (5! / (3! * (5 - 3)!)) = 10
Similarly, the number of ways to choose 2 blue balls without replacement can be calculated as:
5C2 = (5! / (2! * (5 - 2)!)) = 10
Next, we need to multiply these two combinations together:
(5C3) * (5C2) = 10 * 10 = 100
Now, we need to consider the number of ways to arrange these 5 chosen balls (3 red and 2 blue) in a sample of 5. This can be calculated using the permutation formula:
4P2 = (4! / (4 - 2)!) = 12
Finally, we divide this by the total number of ways we can choose any 5 balls from the box, which we've calculated as 20C5 (15504).
So, the probability of choosing 3 balls of one color and 2 balls of another color is:
(10 * 10 * 12) / 15504 = 1200 / 15504
Simplifying this fraction further will give you the answer.