Question

A fair coin is tossed once. If it turns up heads, a balanced die is
rolled twice. Then the sum of the number of dots on the faces
that show up is recorded. If the coin turns up tails, the same die
is rolled once. In this case the number of dots on the face that
shows up is recorded. What is the probability that the recorded
number is larger than 5?

Answer 1

Answer:

4/9

Step-by-step explanation:

when,balanced die is  rolled twice , total result = 6× 6 = 36

results when sum of dots less or equal 5 =  (1, 1), (1, 2), (1,3),(1, 4), (2, 1), (2, 2),

                                                                        (2, 3), (3, 1), (3, 2),(4, 1)

NO. of results when sum of dots is less or equal 5 = 10

NO. of results when sum of dots  is larger than 5 = 36 - 10 = 26

probabilty of turns up head and sum of dots is larger than 5 = $\frac{1}{2} \times \frac{26}{36}$ =  $\frac{13}{36}$

Again, probabilty of turns up tail and no. of dots is larger than 5 = $\frac{1}{2} \times \frac{1}{6}$ = $\frac{1}{12}$

probability that the recorded  number is larger than 5 = $\frac{13}{36} + \frac{1}{12}$ = $\frac{16}{36}$ = $\frac{4}{9}$

Answer 2

Answer:

when,balanced die is  rolled twice , total result = 6× 6 = 36

results when sum of dots less or equal 5 =  (1, 1), (1, 2), (1,3),(1, 4), (2, 1), (2, 2),

                                                                       (2, 3), (3, 1), (3, 2),(4, 1)

NO. of results when sum of dots is less or equal 5 = 10

NO. of results when sum of dots  is larger than 5 = 36 - 10 = 26

probabilty of turns up head and sum of dots is larger than 5 = =  

Again, probabilty of turns up tail and no. of dots is larger than 5 = =

probability that the recorded  number is larger than 5 = = =

Step-by-step explanation: