Are A and B independent events?

2 fair and regular dice are rolled. Let's define the following 2 events.


A - event that the first die is a 4


B - event that the sum of the two dice is 7





Are A and B independent or dependent?

Are A and B independent events?
They are independent, since P(B|A) = P(B) = 1/6.





(Assuming these are 6-sided dice, of course! Otherwise it's dependent.)


§





*** TO STEINER:





Oops. :-)





For any statement A of the form "the event that the first die is k" (for 1≤k≤6), A and B are independent.





On the other hand, if A were as above and B were any sum other than 7, then the events would be dependent.
Reply:Well, though this doesn't sound natural, from the mathematical standpoint they are independent, because P(B | A) = P(the second die is 3) = 1/6 and





P(B) = P(1,6) + P(2,5) + P(3,4) + P(4,3) + P(5,2) + P(6,1) = 6/36 = 1/6





So, P(B | A) = P(B) and, from the definition of independent events, it follows A and B are independent.





But if you defined A = {event that the first die is a 1} and B as before, they wouldn't be independent.
Reply:A %26amp; B are not independent of each other unless P(A)P(B) = P(A%26amp;B). Find out if that holds here.
Reply:P(A) = 1/6 (edited from earlier figure 4/6)


P(B) = 6/36 = 1/6


(because 6 outcomes of a total of 36 are favourable to the sum 7)


A∩B is the event that the first dice has 4 and the sum is 7 which can happen only in one way.


=%26gt; P(A∩B) = 1/36


P(A) * P(B) = (1/6) * (1/6) = 1/36


As P(A∩B) = P(A) * P(B), Aand B are independent.


NOTE: I saw starwhitedwarf's remarks about my error after I edited it. Thanks anyway. starwhitedwarf's answer is perfect.
Reply:the answer is independant as the answer(7) depends on the value got in the second die ,but the value of the first die is already fixed so the event is an independent event
Reply:OMG, what pluralism in the answers so far! It's useful in the politics, but inadmissible in mathematics, so I decided to answer also.


Let's follow the Deep B's advice to find out if


P(A) * P(B) = P(A ∩ B) holds here.





All possible cases rolling 2 FAIR DISTINGUISHABLE CUBIC dice are summarized in the table:


11 12 13 14 15 16 . /second diagonal - favorable for B/ * * * *


. . . . . . . . . .25 . .


. . . . . . . 34. . . . . .


41 42 43 44 45 46 . /4th row - favorable for A/


. . 52 . . . . . . . . . . .


61 . . . . . . . . . . . . . . /first digit of each pair - 1st dice/





Now P(A) = 6/36, P(B) = 6/36 and P(A ∩ B) = 1/36 ("43" above is the only favorable case in the intersection of the 4tn row and the 2nd diagonal). Having (6/36)*(6/36) = 1/36 we must conclude that A and B are INDEPENDENT. Agree everybody?





Now let's roll 2 TETRAHEDRAL dice, faces marked with 2, 3, 4, and 5 (why not, the author has said "fair and regular" only!). The table is:


22 23 24 25 /diagonal left-down favorable for B/


32 33 34 35


42 43 44 45 /this row favorable for A/


52 53 54 55


Now P(A) = P(B) = 4/16 and P(A ∩ B) = 1/16 = P(A)*P(B), so A and B are independent again! (I allowed myself a little joke with jeredwm, regarding his note about 6-sided dice).





Octahedral dice will need markings 0, 1, . . , 7 instead of 1, 2, . . , 8 and the same will work. I already feel much tired to examine the last 2 Platonian solids.





EDIT: STEINER, what have You written about redefining A with the 1, for God's sake? Look at * * * * above, what's the difference?
Reply:dependent.. because when event A occurs, B can also be occur:





eg: 4 and 3 = 7
Reply:P( B) = 6/ 36 = 1/ 6 { (3,4) ; (2, 5); (1, 6) and swap them around}


P (B/A) = 1 / 6


yes the 2 events are independent.





I disagree with the respondent above. P( A) = 1/6 { only one way the first die lands on a 4}, hence this result also gives us the events are independent, since P(A %26amp; B) = 1/ 36 = 1/6 * 1/6 = P(A) * P(B)