Question

Find the conditional probability, in a single roll of two fair 6-sided dice, that neither die is a three, given that the sum is greater than 6 7 The probability is 12 (Type an integer or a simplified fraction)

Answer 1

tial. Tatal autcomes = 36 The sum is greater than 6 possible cautcomes is - {(1,6),(2,5),(2,6), (3;4),(3,5), (3,0),(4,3),(4,4), (41,5),(4,6),(5.2'),(5,3),(5,4) 15,5),(5,6),(6, 1), (6,2), (6,3), 16,4),(6,5), (6,7) } sum is y6 21 2 out of 21 possible outcome the no. of of possible autcomes for neither die is a three is {(1,6), (2,5),(2,6), (4;4),(4;5),(4,6),(5, 2), (5,4),(5,5),(5,6),(6,4),(6,2), (6,4), (6,5),(6,4)} is outcomes 6 j ) ) = P ( Neither three l greater than 6) 5 - sl 15 21 이 가 7 11 0.71428571429 0.7143 Round 4 decimal 22

Answer 2

SOLUTION :

Number of events that neither die is a 3, are :

= 5 outcome on one die * 5 outcomes on o]the other die 

= 25 events. 

Number of events having sum > 6 

Sum 7 :  (1, 6), (6,1), (2, 5), (5, 2), (3, 4), (4, 3) : 6 events

Sum 8 : (2, 6). (6, 2), (3, 5), (5, 3), (4, 4)            : 5 events

Sum 9 : (3, 6), (6, 3), (4, 5), (5, 4)                      : 4 events

Sum 10 : (4, 6), (6, 4), (5, 5)                               : 3 events

Sum 11 : (5, 6), (6, 5)                                         : 2 events

Sum 12 : (6, 6)                                                    : 1 event

                                                                           __________

Total                                                                    : 21 events

                                                                           ___________ 

Events common to both cases  : Above 21 events - Events  (3, 4). (4, 3), (3, 5), (5, 3), (3, 6), (6, 3) :

=> Total such events = (21 - 6) = 15 events

P( Neither die is 3 | sum > 6) 

= Number of events common to both cases / Space of sum > 6

= 15 / 21 

= 5 / 7 (ANSWER).