SOLUTION: Consider rolling a fair 6-faced die twice. Let A be the event that the product of the two rolls is at most 5, and B be the event that the first one is a multiple of 3. (a) What i

Question 1087651: Consider rolling a fair 6-faced die twice. Let A be the event that the product of the two rolls is
at most 5, and B be the event that the first one is a multiple of 3.
(a) What is the probability that the product of the two rolls is at most 5 given that the first one is a
multiple of 3? Show all work. Just the answer, without supporting work, will receive no credit.
(b) Are event A and event B independent? Explain.

Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
Consider rolling a fair 6-faced die twice. Let A be the event that the product of the two rolls is
at most 5, and B be the event that the first one is a multiple of 3.
(a) What is the probability that the product of the two rolls is at most 5 given that the first one is a multiple of 3?
(b) Are event A and event B independent? Explain.

Solution:
Events:
A: product of the rolls is at most 5
B: the first roll is a multiple of 3 (which includes 3 itself).
then
The sample space of A is {(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(3,1),(4,1),(5,1)}, so there are 10 outcomes out of 36, so P(A)=10/36=5/18
The sample space of B is {3,6}, or 2 outcomes out of 6, so P(B)=1/3

(a) P(A|B)=P(A∩B)/P(B)=(1/36)/(1/3)=3/36=1/12
Check: if first roll is a three, then second roll must be a 1 to succeed (1/6)
if first roll is a six, A cannot happen, so (0/6). The total probability is therefore (1/6)(1/2)+(0/6)(1/2) = 1/12.

(b) If A and B are independent, P(A)=P(A|B)
Since P(A)=(5/18)=5/18, and P(A|B)=1/12 which is different from P(A), we conclude that A and B are not independent.

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