Probability for two different events

Question

Let's say I'm playing a game with my friend that involves rolling dice.
The condition for victory is that i must roll a dice and if i roll a 1, i win.
However if i fail the first time i still have a second chance and if i roll again and get either a 1 or 2 this time,i still win. How do i find the probability of being successful?

Answer 1

This problem is may be more easily done by figuring out the probability that you won't win.  Once you know that, you can subtract it from 1 to get the probability that you WILL win.

 

So in order to lose, you have to do the following:

 

NOT roll a 1 on the first roll.  That is, roll a 2, 3, 4, 5, or 6.  The probability of this is 5/6, since there are 5 ways to fail on the first roll, out of 6 possible rolls.

 

Then, you also have to not roll a 1 or 2 on the second roll.  For the same reason the probability of failing on the second roll is 4/6, or 2/3.

 

Since the two rolls are independent, the probability of failing on both rolls is the product of the two individual probabilities:  5/6 * 2/3 = 10/18 = 5/9.

 

So the probability of losing is 5/9.  Therefore the probability of winning is 1 - 5/9 = 4/9.