# Probability homework question

If there are 6 red and 30 white beads. What are the chances of drawing red in 2 successive trials ?

2.If red bead is not replaced.

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

• Actually I am confused here. Do we multiply these probabilities or add them ?
– sss
Oct 3, 2013 at 19:27
• @sss: $\frac6{36}+\frac5{35}$ is bigger than $\frac6{36}$, and $\frac6{36}\cdot\frac5{35}$ is smaller than $\frac6{36}$. Which is more likely, drawing just once and getting a red ball, or drawing twice and getting two red balls? Oct 3, 2013 at 19:33
• Here it is multiply as two events should occur simultaneously Oct 3, 2013 at 19:34
• @BrianM.Scott Drawing once and getting a red ball.
– sss
Oct 3, 2013 at 19:35
• @saikirangrandhi Thanks a lot.
– sss
Oct 3, 2013 at 19:36

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.