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

Re: basic conditinal prob.

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

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: $\displaystyle \frac{\frac{1}{12}}{\frac{1}{12}+\frac{5}{12}+ \frac{1}{12}}$

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

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.