Let's say you are playing a game of chance. To win the game an event has to occur. That event only has a 10% to occur. You can play the game 10 times. How should we calculate the probability of winning the game
- At least once
- All ten times
- Zero times.
For the $2^{nd}$ one, I know I have to add probabilities if events are independent of each other. But when I add the probabilities of winning ten times, i.e, $\frac{1}{10} + \frac{1}{10} +\frac{1}{10} +\frac{1}{10} +\frac{1}{10} +\frac{1}{10} +\frac{1}{10} +\frac{1}{10} +\frac{1}{10} +\frac{1}{10} = 10*\frac{1}{10} = 1$. Which obviously isn't true. I'm still new to probability, so did I do something wrong for the second one.
Best Answer
A good "rule of thumb" with probability is that when you have "and", you multiply, and when you have "or", you add.
So in this case you want to win game 1 $\textbf{and}$ game 2 and so on, which means your computation is: $$\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}=\frac{1}{10^{10}}$$