# Probability

• September 7th 2011, 11:47 AM
calculuskid1
Probability
Consider the trial on which 3 is first observed in successive rolls of a six-sided die. Let A be the event that 3 is observed on the first trial. Let B be the event that at least two trials are required to observe a 3. Assuming that each side has a probability of 1/6, find (a) P(A), (b) P(B),(c) P(A U B).
so (a) im guessing is just P(A)=1/6
(b) is at least two rolls? does that mean 1/12?
(c) P(A U B)= 1/6 + 1/12 - 0
Can some one please correct me if I am wrong?
• September 7th 2011, 11:54 AM
calculuskid1
Re: Probability
after carefully re-evaluating, b) should be P(B)=1-P(B')=5/6
• September 7th 2011, 01:06 PM
Plato
Re: Probability
Quote:

Originally Posted by calculuskid1
after carefully re-evaluating, b) should be P(B)=1-P(B')=5/6

That is now correct.
Note that events $A~\&~B$ are mutually exclusive so what is $P(A\cup B)~?$