# Probability problem

1. Sep 14, 2008

### alef

1. The problem statement, all variables and given/known data

we have 2 boxes with black and white balls inside (box#1 = white balls, box#2 = black).all black and white balls are numbered from 1 to 5.
person take out 3 balls from the 1st box and 3 balls from the 2nd box. if success event is when we get first three balls what the probability of success in at least one of the two tasks merely by chance if the choices were made randomly and independently?

2. The attempt at a solution
as for me, we have equal probability to get right set of balls (1 + 2 + 3): (1/5)(1/4)(1/3). therefore the probability of success in at least one of the two tasks should be equal two over the number of all possible variations to get 3 balls from 5 in specific order?
but, in fact, this reasoning doesn't give a right answer. please fix my attempt to solve this problem.

2. Sep 14, 2008

### Redbelly98

Staff Emeritus
For drawing from one box, what is:

A. The probability that the 1st ball taken is either #1, #2, or #3?
B. The probability that the 2nd ball taken is one of the remaining two balls we need?
C. The probability that the 3rd ball taken is the remaining ball needed?

Multiply those three answers together, and you'll have the probability that ball #'s 1, 2, and 3 were drawn from a single box.

p.s. Welcome to Physics Forums!

3. Sep 14, 2008

### arunbg

First let me clarify the question.

There are 2 bags with 5 balls each. Bag 1 contains all white balls numbered from 1 to 5, and bag 2 contains all black balls numbered from 1 to 5. 3 balls each are drawn at random from the two bags. What is the probability of getting balls numbered 1,2,3 for at least one color?

If this is the question, you need to start by calculating the probability of drawing balls numbered 1,2,3 from one bag. Order is irrelevant here I believe, so you can use combinations to find that. Once you get this probability, subtracting it from 1 gives you the probability of the complementary event ie no such set 1, 2, 3 . Assuming that the draws from each bag are independent of each other,
P(at least 1 such set) = 1 - ( P(no such set from a bag) )^2
Can you finish?

4. Sep 14, 2008

### alef

not exactly.. you are right - we have 2 bags with 5 balls each. Bag 1 contains all white balls numbered from 1 to 5, and bag 2 contains all black balls numbered from 1 to 5. in first trial we take out only a ball from, say box 1.in the second trial we take out only a ball from the box #1 also. in the third trial we take out the third ball from the box #1. our experiment will be successful only in case when we achieve 1-2-3 order. the same with the box #2.

lets look only at the box1 at first.
all possible orders is 5P3=60 different outcomes. successful only 1 when we have 1-2-3.
the same with the box2.
but i can't understand where we can apply condition "the probability of success in at least one of the two tasks merely by chance"?

p.s. thanks for greeting me here:) I want to believe I can help someone to

5. Sep 14, 2008

### Redbelly98

Staff Emeritus
That's good so far. And how many different ways could we have drawn balls 1, 2, and 3?

6. Sep 14, 2008

### alef

one for each box. total = 2

7. Sep 14, 2008

### Redbelly98

Staff Emeritus
No, remember we're just looking at box 1 for now.

For just 1 box:

We might draw ball #1 first, #2 second, and #3 third.
or
#3 drawn first, #1 second, #2 third,
etc. etc.

8. Sep 14, 2008

### alef

but #3+#2+#1 is not the same as #1+#2+#3

9. Sep 14, 2008

### Redbelly98

Staff Emeritus
Yes, but wouldn't it still count as a "success" in that we drew the first 3 balls?