Transition Algebra 1 I need help on my Credit Recovery stuff at school THANKS!!!

Question

Eden is tossing 2 number cubes with faces numbered from 1 to 6. To win the game, she must toss an even number on the first cube and an odd number on the second cube. What is the probability Eden will toss an even number on the first cube and an odd number on the second cube?

Answer 1

if you roll the die twice, it will have total of 36 outcomes. where 6 choices on the first roll and 6 choices on the 2nd roll, which 6x6 = 36

 

for your question, there is 3 outcomes if you roll for a even number for the first time so, the probability will be 3/6 or 1/2.

for the second time roll, it will also be 3/6 or 1/2 for odd number.

 

due to the two outcomes are independent. the probability for both rolls will be

(1/2)*(1/2) = 1/4

 

 

Another way of seeing this.

 

below are the 36 possibilities.  9 of them is the one that satisfy your requirement. 9/36 = 1/4 

 

(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)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)