QUESTION 8The numbers showing on the upper faces of two, fair, six-sided dice are observed when rolled.  Consider events A and B defined as:A = the sum of the two values observed on the upper faces is at least 8 B = at least one dice has a 2 or less on the upper faceFind 1. P(B')    (3 decimal places)  2. P(A or B)    (3 decimal places)

1. P(B')

Event B is defined as "at least one dice has a 2 or less on the upper face". The complement of this event, B', is "neither dice has a 2 or less on the upper face".

There are 4 outcomes (3, 4, 5, 6) out of 6 that satisfy B' for a single die. Since the two dice are independent, we multiply the probabilities:

P(B') = (4/6) * (4/6) = 16/36 = 0.444 (to three decimal places)

2. P(A or B)

Event A is defined as "the sum of the two values observed on the upper faces is at least 8". There are 15 outcomes out of 36 that satisfy this condition: (2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6). So, P(A) = 15/36 = 0.417 (to three decimal places).

Event B is defined as "at least one dice has a 2 or less on the upper face". There are 20 outcomes out of 36 that satisfy this condition: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (4,1), (5,1), (6,1), (3,2), (4,2), (5,2), (6,2). So, P(B) = 20/36 = 0.556 (to three decimal places).

However, there are 5 outcomes that satisfy both A and B: (2,6), (3,5), (4,4), (5,3), (6,2). So, P(A and B) = 5/36 = 0.139 (to three decimal places).

The probability of A or B is given by P(A) + P(B) - P(A and B):

P(A or B) = 0.417 + 0.556 - 0.139 = 0.834 (to three decimal places).