Abox contains five slips of paper. each slip has one of the number 4, 6, 7, 8, or 9 written on it and all numbers are used. the first player reaches into the box and draws two slips and adds the two numbers. if the sum is even, the player wins. if the sum is odd, the player loses.

70% chance of winning

Step-by-step explanation: What is your question? Are you trying to ask the probability of winning? (I will assume this and answer)

The whole case of selecting to numbers : 5C2 = 5 X 4 / 2 = 10

the cases of getting a odd sum : select 1 odd number and 1 even number

=> select 1 odd number : 7 or 9 => 2 cases

=> select 1 even number: 4,6,8 => 3 cases

you multiply 2 and 3 and divide it by 2 because order doesn't matter

so the answer is 1 - (3/10) = 0.7  

Hi

A box contains five slips of paper.  

Each with the numbers 4,6,7,8,or 9 written on it and all of the numbers are used

Two slips drawn

P(sum is even and player wins) =  13/25  

___4_ 6_ 7__ 8__ 9

4|_8_10_11__12__13

6|_10_12_13__14__15

7|_11_13_14__15__16

8|_12_14_15__16__17

9|_13_15_16__17__18

  3E 3E 2E  3E  2E

Does the probability change if the two numbers are multiplied?

Yes, multiplying the number results in many more even results.  

Step-by-step explanation: