Basic probability…balls extraction im stuck with a basic probability problem:
n balls --> n-1 extraction. Only one black ball. No replacement.
example:
7 balls(6 white, 1 black). 6 extractions.
i know that the probablity of 6 whites is: 6/7 · 5/6 · 4/5 · 3/4 · 2/3 · 1/2 = 0.14 aprox, so the prob of get the black ball is: 1 - 0.14 = 0.86
but how can i calculate the probabilities of black in six extraction without using the above technique?
Regards
 A: Another thinking of your answer would be $1-{1\over{7\choose 6}}$ but is essentially the same thing as the above technique.
Other than that you would use ${6\choose 5}\over{7\choose 6}$ by fixing the black ball to be chosen and choose $5$ from the remaining $6$ balls.
A: I think you're meant to realise that the probability of drawing $n-1$ balls is the same as the probability of leaving $1$ ball in the bag at the end. Therefore probability of six white draws $= \frac{1}{7}$ and probability of five white and one black is the same as leaving a white behind $=\frac{6}{7}$
A: Hint:
Extracting $n-1$ balls from $n$ balls without replacement (1) comes to the same as extracting $1$ ball from $n$ balls without replacement (2). Just think of this extracted ball in (2) as the unique ball wich is not extracted in (1).
A: The probability of drawing a black ball is the complement of not drawing a black ball:
$$P (\mbox {drawing black})=1-P (\mbox {not drawing black})$$
Not drawing the single black ball when drawing $n-1$ balls from $n$ balls is simply $\frac {1}{n} $, so the probability you're after is:
$$1-\frac {1}{n}=\frac {n-1}{n}$$
A: If you have $n$ balls, you can, in the beginning, fix the black ball (i.e. it'll be present in all of your combinations). Then, you can choose $n-2$ balls from the remaining $n-1$ balls. You can do it in ${n-1\choose n-2}=n-1$. If you were to choose any $n-1$ balls from $n$ balls, you could do it in ${n \choose n-1}=n$ ways. So, the probability of getting the black ball in your combination is $\frac{n-1}{n}$.
A: *

*You want to draw 6 balls and the remaining ball should be black. 

*Exactly one of your seven balls is black. So the six balls you draw in 1. are  white and the remaining ball is black

*If you draw 7 balls (all balls) then 2. is the same as if you draw 6 white balls and at last a black ball. So the balls are drawn in the following order: w w w w w w b

*The sequence does not matter. 3. has the same probability as drawing a black ball with the first draw and drawing white balls in the remaining 6 draws. Here the balls are drawn in the following order: b w w w w w w
5.Because there is only one black balls. is the same as drawing one ball, which is black. The remaining balls are all white.

*The probability of drawing a black ball is $\frac{1}{7}$

