Question

probabilities Find the probabilities for each event. Consider rolling a pair of fair dice two times. Let A be the total on the up-faces for the first roll and let B be the total on the up-faces for the second roll.

1. A = {2 on the first roll}, B = {8 or more on the first roll}.
2. A = {2 on the first roll}, B = {8 or more on the second roll}.
3. A = {5 or less on the second roll}, B = {4 or less on the first roll}.
4. A = {5 or less on the second roll}, B = {4 or less on the second roll}.

Answer 1

GIVEN THAT

(a) Events A and B are disjoint, because they have no common outcomes. P(combination of the events)= 0, because combination of these two events is not possible.

(b)Events A and B are independent, because the occurrence of one event does not effect the occurrence of the another event. Here P(combination of the events)=P(A).P(B). Now P(A)=1/36, because 2 on the first roll is only possible when 1 is on one dice and 1 is also on another. P(B)=P(getting 8 or more)=P(getting 8 or 9 or10 or 11 or 12)=P(8)+P(9)+P(10)+P(11)+P(12), since they are all mutually exclusive events

P(getting 8on second roll)=co-efficient of x6 in (1-x6)2.(1-x)-2/62, using De Moivre's theorem,

=5/36

similarly, P(9)=4/36, P(10)=3/36, P(11)=2/36, P(12)=1/36

P(getting 8 or more)=15/36 and P(combination of the events)=P(A).P(B)=1/36.15/36=5/432

(c)Events A and B are independent. Here P(combination of the events)=P(A).P(B). Now P(A)=P(5)+P(4)+P(3)+P(2)=4/36+3/36+2/36+1/36 (using De Moivre's theorem)=10/36. Similarly P(B)=6/36. Therefore, P(combination of the events)=5/108

(d)Events A and B are neither independent nor disjoint.Here P(combination of the events)=P(getting 4 or less on the second roll)=6/36=1/6