Probability homework question If there are 6 red and 30 white beads. What are the chances of drawing red in 2 successive trials ? 
1.If red bead is replaced.
2.If red bead is not replaced.
 A: Prob of choosing a first red bead = 6/36
Now it is not replaced we have 5 red beads and 35 in total.
If it is replaced you have the situation as before taking the first bead
I think you can continue from here
Edit: Multiply is the case when the events should occur simultaneously, Addition is when something like anyone of the events can occur
A: I am assuming you mean successive as in sequential and not as in success.
$$P(red bead, 1st trial) = 6/36 = 0.1667$$
$$P(red bead, 2nd trial) = 5/35 = 0.1429$$
These are approximations to four decimal places. This is also assuming the bead is not replaced after each trial. If they are replaced:
$$P(red bead, 1st trial) = 6/36 = 0.1667$$
$$P(red bead, 2nd trial) = 6/36 = 0.1667$$
Since you're looking for $P(red on 1st and red on 2nd)$, you multiply the probabilities:
$$P(if bead is not replaced) = (6/36)*(5/35) = 0.0238$$
$$P(if bead is replaced) = (6/36)*(6/36) = 0.0278$$
Hope this helps. Remember that for and logic within probability it is multiplied, and for or logic it is added.
