8 blue balls, 4 red balls. Remove 3 balls. probability for remove 3 blue balls?

Question

a.0,1817 b.1,1818 c.0,1819 d.0,1820

Answer 1

25.5 % chance

2/3 for the first ball
7/11 for the next ball
3/5 for the last ball
multiply the numerators then denominators and you get:
42/165 = .2545454545

Answer 2

The probability of removing 1 blue ball is 2/3, and if the first one is blue, the second one being blue is 7/11, and the third one being blue is 3/5.

2/3 * 7/11 * 3/5
= 2/1 * 7/11 * 1/5
= 14/55

None of the above is the answer, it's 14/55!

Answer 3

well you have a total of 12 balls and you want to remove 3 blue so 3/12 = 1/4

Answer 4

there's 12 balls. the probability of picking blue out of 12 balls is 8/12 the first time. to pick another blue out of the remaining 11 is 7/11. the prob of picking a third blue out of the remaining 10 is 6/10.

prob1B(8/12)prob2B(7/11)prob3B(6/10) = .2545

Answer 5

I've seen a similar question before. And like it, this one fails to specify everything needed to work the problem. For example, is there replacement after each of the three draws? Or does the number of balls available diminish by one each draw?

Finally, why are you asking for an answer without showing the work to get that answer? Do you want to understand probability or just work a homework problem?

Answer 6

the probability for the first blue ball=8/12;secon=7/11 and the third=6/10 and so the probability for the three blue balls will be
(8/12)(7/11)(6/10)=0.25454545........................

Answer 7

the probability is 1 / 4

Answer 8

You should ask on how to do the question, not for the answer. Go read your text book.

Answer 9

8C3/12C3=56/220=14/55= 0.254545454

Answer 10

b