probability of 2 events happening I am just starting probability in school and would like some help please with below questions:
The work i have so far is this:
(i) 1/8 * 1/8 = 1/64   although it says "in that order" and this is confusing me
(ii) 1/8 *1/8 = 1/64   i am assuming this is correct also?
(iii) 1/8 * 1/8 * 1/8 * 1/8 * 1/8 = 1/32768   as there is 4 odd numbers: (1,3,5,7) and also it asks to use 6
(iv) 1/8 * 1/8 * 1/8 * 1/8 * 1/8 = 1/32768 as the prime numbers are: (2,3,5,7) and also it asks to use 2
I would appreciate some help as i am only learning this new subject. thank you

 A: $1)$ For first part you are right. In that order means $2$ has appeared first time, and after that, $4$ has been appeared. If it was not mentioned "in that order", then the probability was $2\times \frac{1}{8}\times \frac{1}{8}$
$2)$ You are right
$3)$ You are wrong. There are $4$ odd numbers $\{1,3,5,7\}$, then it will be $\frac{4}{8}\times\frac{1}{8}\times2$, in which $2$ has been written, since it has not asked that the outputs to be in order
$4)$ You are wrong. There are four prime numbers: $\{2,3,5,7\}$, thus the intended probability is $(\frac{1}{8}\times\frac{1}{8})+(\frac{1}{8}\times\frac{3}{8}\times2)$, which has been obtained in the following manner. As $2$ is a prime number itself, consider it separately. When there are 2 at both moves, the probability is $(\frac{1}{8}\times\frac{1}{8})$, and when $2$ has not appeared as a prime number, there are 3 prime numbers left as a choice. 
A: (i) This is correct. To ease your confusion, maybe you should ask yourself what is the probability of choosing 2 and 4, in any order.

 Since the only possibilities are choosing 2 then 4, or 4 then 2, the answer is two times the answer you gave.

(ii) Also correct.
(iii), (iv) These are not correct. First, ask yourself the following simpler questions:


*

*If you pick one disk, what is the probability that it is odd?

*If you pick one disk, what is the probability that it is prime?



 The answer to both these questions is $\frac{1}{8}+\frac{1}{8} + \frac{1}{8} + \frac{1}{8}=\frac{1}{2}$, not $\frac{1}{8}\cdot\frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{8}$. Do you see why?

For (iii), you also have to consider whether you pick 6 first, or the odd first.
For (iv), you have to consider not only the order, but also the fact that 2 is a prime! For example, picking 2 first, and then picking 2 again is valid.

If all else fails, enumerate all possible outcomes (there are $8 \cdot 8 = 64$ of them), count how many of these outcomes satisfy the condition, and then divide by $64$.
