How Many Ways Can You Choose Two Balls From a Set of Black and White Balls?

[jack1234]

There are 6 white and 5 black ball. How many ways to choose 2 balls from it?
My Answer:
3 ways, which is
(1)two white ball
(2)One white and One Black
(3)Two Black ball

What is the probability that one of the drawn balls is white and the other black?
My Answer:
C(6,1)*C(5,1)/C(11,2)

If my answer for this two question is correct, can I conclude that, for combination, we view all black ball as identical,
but for the sample space of probability(if we view it as unordered set of drawn balls) we treat all the black balls as distinct, hence we have C(11,2)?

 

[e(ho0n3]

Regarding your first question, there are three possible outcomes which you correctly mention. However, the number of ways to choose the balls is not the same as the number of possible outcomes.

The answer to your second question is correct.

The comment at the end makes no sense. C(11,2) represents the number of ways of picking 2 objects from a set of 11 of them regardless of what they are (hence no distinction is made).

 

[HallsofIvy]

"First white and second black" and "first black and second white" are distinct ways that give one white and one black ball.

There four ways of choosing two balls and two of them give "one black and one white".

 

[jack1234]

Thanks for the replies :) However, I have some question regarding the replies.

Regarding your first question, there are three possible outcomes which you correctly mention. However, the number of ways to choose the balls is not the same as the number of possible outcomes.

The answer to your second question is correct.

The comment at the end makes no sense. C(11,2) represents the number of ways of picking 2 objects from a set of 11 of them regardless of what they are (hence no distinction is made).

Do you mean that first question is wrong, because the way of choosing it is C(11,2), but the number of distinct combination(outcome) is 3?

"First white and second black" and "first black and second white" are distinct ways that give one white and one black ball.

There four ways of choosing two balls and two of them give "one black and one white".

For the first question, I am just choosing it(combination), but not going to order it(permutation), hence isn't the number of distinct outcome be 3; and the way of choosing it be C(11,2) as mentioned by e(ho0n3?

 

[HallsofIvy]

To second question: No. The four ways of choosing 2 balls are "white, white", "black, white", "white, black", and "black, black". Since there are six white and 5 black balls, the probability of drawing a white ball on the first draw is 6/11. Then there are 5 white and 5 black left so the probability of drawing a white ball on the second draw is 5/10= 1/2. The probability of "white, white" is (6/11)(1/2)= 3/11.

The probability of drawing a black ball on the first draw is 5/11 and then there are 6 white and 4 black balls left. The probability of drawing a white ball now is 6/10= 3/5. The probability of "black, white" is (5/11)(3/5)= 3/11. I'll let you calculate the probailities of the other two cases and add them.