The events A and B are independent therefore P(A and B) = P(A)P(B) = (1/2)(1/6) =1/12
P(AUB) = P(A)+P(B)-P(A and B)
The conditional robability P(X/Y) =(P(X and y))/P(Y) JUST APPLY THE FORMULA..ITS EASY
You flip a coin and a die once
Let A = head and let B = get a six
You run the experiment and you are told that at least one event occurred.
What is the probability that both A and B occurred?
at least = 7/12
But dont know how to calculate P(AB | A u B)
Definition of independent event:
Independent events are events which have no effect on one another.
In your case how the result of the coin will have any effect on the result of the dice?
This is wrong ....
I suggest you to consult a good book on probabilities ( STATISTICS1 STEVE DOBBS & JANE MILLER PAGE 76 EXAMPLE 4.5.1 )
The coin can be T or H, the die can be 1, 2, 3, 4, 5 or 6, each of those being "equally likely".
So, when a coin is flipped and a die is rolled, the "equally likely" events are
You are told that "at least one" of "Heads" and "6" occured so that reduces to
a total of 7 equally likely events. "Both A and B", that is, a Head and a 6, is one of those 7 events: probability of "head and 6" given that at least one of "head" and "6" occured is 1/7.