Simple Stats Q on dependent/independent events?

Question

Toss 2 fair dice. There are 36 outcomes each with probability 1/36.
Let:
A be the event '4 on first die'
B be the event 'sum of numbers is 7'
C be the event 'sum of numbers is 8'

Show that A and B are independent.
Show that A and C are dependent.


I missed a lecture on this. Can someone show me how to do this? do you check that P(B/A) = P(B)? or something?

why are (4,1) (4,2) favourable? and not (4,3) (4,4).....?

you haven't explained the independency thing

Answer 1

P(A) = 1/6
B = (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
P(B) = 6/36 = 1/6
A and B = (4,3)
P(A and B) = 1/36
P(A)P(B) = (1/6)(1/6) = 1/36 = P(A and B)
Therefore A an B are independent events.

C = (2,6), (3,5), (4,4), (5,3), (6,2)
P(C) = 5/36
A and C = (4,4)
P(A and C) = 1/36
P(A) P(C) = (1/6)(5/36) = 5/216
P(A)P(C) is not equal to P(A and C)
Therefore A and C are not independent

NOTE: The equation P(B|A) = P(B) and the equation P(A and B) = P(A)P(B) are equivalent.

Answer 2

A.If the event 4 is on first die, the favorable cases are (4,1) and 4,2) and hence the total number of favorable cases to the event = 2. Therefore the required probability = 2/36=1/18.
B. sum of the numbers 7 can happen in(1,6), (2,5), (3,4), (6,1), (2,5), (3,4) ways. Hence the total number of favorable cases to the event 'sum numbers 7' =6.
Therefore, the required probability = 6/36=1/6.
C. Sum of the numbers 8 can happen in (2,6), 3,5), (4,4) and (4,4). The total number of favorable ways = 4
Hence the required probability = 4/36 = 1/9.