The numerial key for a specific safe is four digits long. The digits must be numbers from $1$ to $6$. (an example will be 4321 or 2564)
a) I threw a die randomly $4$ times. What will be the probability that I open the safe? How about repeating the same process $3$ times?
At a school in Utah there is a bench that can fit 2 people. We took an exam and concluded that it is impossible to copy from someone that is not sitting in the same bench as you. So, the teacher gave assigned sits which were at random. The class consists of twenty students and ten benches. We have that there are 12 A's and 8 B's. Now we have that 8 students are at the same bench where both have A’s and 6 students that has B's sitting at the same bench. What is the probability that this will occur?
My attempt at the solution:
A die contains 6 digits and each digit has a $\frac{1}{6}$ chance of occuring so there is a possible of $6^4$ ways of rolling the dice four times. To get a possible of just one way of arrangments of tosses is just 1. So the probaility of getting the correct code to open the safe then will be $\frac{1}{6^4}$. If we repeat the process three times then it will be $3\dot\ \frac{1}{6^4}$. Is that right?
For the second question I do not know what to do because I don't even understand the question.