Problem:
An urn contains 3 red and 7 black balls. Player A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn)
I have the solution:
$$\Bbb{P}(A)=\frac{3\cdot 9!+6\cdot 3 \cdot 7!+ 7\cdot6\cdot5\cdot4\cdot3\cdot5!+7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot3\cdot3!}{10!}$$
But i have no idea how they came up with this. Can someone explain this? Thank you.
Best Answer
Imagine that the balls have secret identifying numbers. and imagine that we draw out all the balls, whether someone has already won or not.
There are $10!$ ways to draw out the balls, all equally likely.
Player A could win on the first draw. How many sequences correspond to this? The winning red ball could be any of the $3$, and then the rest of the balls could come in any one of $9!$ possible orders, for a total of $(3)(9!)$.
Or else Player A could win on the third draw. That happens if the first ball is black, the second is black, the third is red, and the rest are arranged in any order. The first ball can then be chosen in $7$ ways, and for each choice the second can be chosen in $6$ ways. Then we have $3$ ways for the third, since it has to be red, and then $7!$ for the rest, for a total of $(7)(6)(3)(7!)$. This should be the second entry in the numerator. It isn't, presumably a typo. No wonder you are puzzled.
Or else Player $A$ could win on the fifth draw. The same reasoning gives $(7)(6)(5)(4)(3)(5!)$ ways. That's correctly written down.
Finally, there could be a win on the $7$-th draw. The number of sequences that give this is correctly written down.