A jar contains 7 red balls, 5 green balls, 4 blue balls, and 3 white balls. A sample of size 7 balls is selected at random without replacement . Find t the probability that sample contains two red balls, 2 green balls, 2 blue balls, and 1 white ball.
Let's simplify the problem to having
7 red balls,
5 green balls, and
4 blue balls
(for a total of 16).
Find the probability of drawing
2 red balls, 2R
2 green balls, 2G
2 blue balls, 2B
(total of 6 balls), then
Below we will use the notation of
(n,r)=combination (n,r)
=n!/((n-r)!r!)
Number of ways of drawing
2R,2G and 2B would be
for 2R: (7,2)=21
for 2g: (5,2)=10
for 2B: (4,2)=6
and number of ways of drawing 6 balls out of 16: (16,6)=8008
So probability
P(2R+2G+2B)=(7,2)(5,2)(4,2)/(16,6)
=21*10*6/8008
=45/286
Extend the idea to the problem in hand, and post for answer checking if you wish.
To find the probability of selecting a sample containing two red balls, two green balls, two blue balls, and one white ball, we need to calculate the favorable outcomes and divide it by the total number of possible outcomes.
First, let's determine the total number of possible outcomes. We have a total of 7 + 5 + 4 + 3 = 19 balls in the jar, so there are 19 balls to choose from for the first selection, 18 for the second, 17 for the third, and so on.
The number of favorable outcomes can be found by multiplying the chosen balls for each color category. For red balls, we have 7C2 (combination of choosing 2 from 7), for green balls, we have 5C2, for blue balls, we have 4C2, and for white balls, we have 3C1.
Using the combination formula nCr = n! / (r!(n-r)!), where n is the total number of elements and r is the number of elements chosen, we can calculate the number of favorable outcomes.
Therefore, the probability of selecting a sample containing two red balls, two green balls, two blue balls, and one white ball is:
(7C2 * 5C2 * 4C2 * 3C1) / (19C7)