SOLUTION: I have 27 balls. 14 are red and 13 blue. I deceded to group the balls into a group of 3s ?
the balls will be choosen randomly.
What is the probablity that the group will have 2 r
Algebra.Com
Question 167212: I have 27 balls. 14 are red and 13 blue. I deceded to group the balls into a group of 3s ?
the balls will be choosen randomly.
What is the probablity that the group will have 2 red and 1 blue.
2)-the probability that the group will have all reds or all blue
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
possible groups of 3 out of 27 __ 27C3 __ 27*26*25/(3*2*1) __ 2925
possible groups of 2 red out of 14 __ 14C2 __ 14*13/(2*1) __ 91
possible groups of 1 blue out of 13 __ 13
probability of 2 red and 1 blue __ 91*13/2925 __ .404 (approx)
possible groups of 3 red out of 14 __ 14C3 __ 14*13*12/(3*2*1) __ 364
probability of 3 red __ 364/2925 __ .124 (approx)
possible groups of 3 blue out of 13 __ 13C3 __ 13*12*11/(3*2*1) __ 286
probability of 3 blue __ 286/2925 __ .098 (approx)
RELATED QUESTIONS
I have 27 balls. 14 are red and 13 blue. I deceded to group the balls into a group of 3s... (answered by scott8148)
I have 27 balls. 14 are red and 13 blue. I deceded to group the balls into a group of 3s... (answered by scott8148,Edwin McCravy)
There are 3 red,4 white,5 purple & 6 blue balls inside the box.If I want to have 4 balls... (answered by edjones)
A box contains 4 red balls, 6 black balls, 8 blue balls and 12 white balls. What... (answered by macston)
There are 3 blue balls, 5 red balls, and 4 white balls in a bag of balls. If a person... (answered by rothauserc)
There are 8 blue balls, 8 yellow balls and 8 red balls of the same type inside a sack.... (answered by ikleyn)
If you are going to be drawing 3 balls from a sack containing 4 red balls, 3 green balls... (answered by ikleyn)
if you have Z number of balls in a box which consists of red, green, white and blue... (answered by Theo)
A bag contains 5 red balls and 5 blue balls. 2 balls are selected at random. Find the... (answered by stanbon)