A hamburger chain's game card has ten squares, each of which has a covering that can be rubbed off to reveal what is pictured beneath. Seven squares show different foods, two square show the same food, and one square has an 'X' - YOU LOSE on it.
The game is played by rubbing the covering off the squares until either you get the pair and win, or get the 'X' and lose.
What is the probability that you will win this game?
i didn't get this question not a single bit.
To win the game you must find the two 'good' squares before the one 'bad' square, out of seven uncovered one at a time. Call the other four squares 'okay'.
Suppose it takes $k$ turns to until you win, where $k$ is some number between $2$ and $6$. Then to have won you must have selected $k-2$ 'okay' squares and $1$ 'good' square out of all the ways to select $k-1$ squares out of all $7$, and then have selected the second 'good' square out of the $8-k$ remaining squares.
So the probability of wining is: $${\begin{align} \mathsf P(W) & = \sum_{k=2}^6 \left(\frac {{4\choose k-2}{2\choose 1}}{{7\choose k-1}}\cdot\frac{1}{8-k}\right) \\[1ex] & = \sum_{k=2}^6\left(\frac{4!\,2!\,(k-1)!\,(8-k)!}{(k-2)!\,(6-k)!\,7!\,(8-k)}\right) \\[1ex] & = \frac{4!\,2!}{7!}\sum_{k=2}^6 (k-1)(7-k) \\[1ex] & = \frac{4!\,2!}{7!}\sum_{k=1}^5 k(6-k) \\[2ex] & \ddots \end{align}}$$
Simplify andevaluate.