could you please tell me , am I correct with my answers in this task ?
slot machine has 3 wheels. Each wheel has 11 stops: a bar and the digits 0,1,2,…,9.
When the handle is pulled, the 3 wheels spin independently before coming to rest.
Find the probablity that the wheels stop on the following positions:
3 bars ,
the same digit on each wheel ,
at least one bar
my answers:
1/1331 ,
10 / 1331 ,
331 / 1331
Best Answer
You are correct on the first one: $(\frac{1}{11})^3$ is indeed the probability, which turns out to be $\frac{1}{1331}$ which is what you got.
You are also correct on the second one, we choose $\frac{10}{11}$ of one wheel since a bar is not a number, then there's a $\frac{1}{11}$ chance for the other two, resulting in a probability of $(\frac{1}{11})^2 \cdot \frac{10}{11}$ which is what you got.
For at least one bar, it's better to calculate the probability we get zero bars (complementary counting), and so there are $10$ numbers not bars out of $11$. $11^3-10^3 = 331$, and that over $11^3$ is $\frac{331}{1331}$, which is also what you got.
You got all the answers correct. Good job!
-FruDe