# basic conditinal prob.

• Mar 6th 2013, 08:39 PM
sfspitfire23
basic conditinal prob.
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)
• Mar 6th 2013, 10:22 PM
MINOANMAN
Re: basic conditinal prob.
Sfspitfire

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
• Mar 7th 2013, 03:59 AM
sfspitfire23
Re: basic conditinal prob.
No, you're incorrect. The answer is 1/7. There's no way that they're independent.
• Mar 7th 2013, 04:34 AM
Plato
Re: basic conditinal prob.
Quote:

Originally Posted by sfspitfire23
No, you're incorrect. The answer is 1/7. There's no way that they're independent.

Apply Bayes Rule: $\displaystyle \frac{\frac{1}{12}}{\frac{1}{12}+\frac{5}{12}+ \frac{1}{12}}$
• Mar 7th 2013, 07:56 AM
MINOANMAN
Re: basic conditinal prob.
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 )

minoas
• Mar 7th 2013, 08:18 AM
HallsofIvy
Re: basic conditinal prob.
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
T, 1
T, 2
T, 3
T, 4
T, 5
T, 6
H, 1
H, 2
H, 3
H, 4
H, 5
H, 6
You are told that "at least one" of "Heads" and "6" occured so that reduces to
H, 1
H, 2
H, 3
H, 4
H, 5
H, 6
T, 6
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.