Question

What is the probability of rolling a number greater than or equal to 8 with the sum of two dice, given that at least one of the dice must show 6?

Answer 1

Answer:

a dice goes from 1-6

6+1=7

6+2=8

6+3=9

6+4=10

6+5=11

6+6=12

Of those possible choices, 5 are greater than or equal to 8. So the answer is 5/6.

mark brainiest please

Answer 2

Answer with explanation:

⇒Total Possible outcome when a dice is thrown twice =6²=36

={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4).......(6,5)}

→Total Favorable Outcome = Sum of the Number on the dice should be greater than or equal to 8 and one of the dice must show 6.

→A =Sum of the two numbers on the dice should be ,8, 9,10,11,12.

={(4,4),(5,3),(3,5),(6,2),(2,6),(6,3),(3,6),(5,4),(4,5),(5,5),(4,6),(6,4),(6,5),(5,6),(6,6)}

=15

→B={(6,2),(2,6),(6,3),(3,6),(4,6),(6,4),(6,5),(5,6),(6,6)}

→A ∩ B=B

Total Favorable Outcome ={(6,2),(2,6),(6,3),(3,6),(4,6),(6,4),(6,5),(5,6),(6,6)}=9

Required Probability

              $=\frac{\text{Total Possible Outcome}}{\text{Total Possible Outcome}}\\\\=\frac{9}{36}\\\\=\frac{1}{4}$