SOLUTION: Seven slips of paper marked with the numbers 1, 2, 3, 4, 5, 6, and 7 are placed in a box and mixed well. Two are drawn. What are the odds in favor of the sum of the numbers on the

Question 289407: Seven slips of paper marked with the numbers 1, 2, 3, 4, 5, 6, and 7 are placed in a box and mixed well. Two are drawn. What are the odds in favor of the sum of the numbers on the two selected slips is not 6?
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
51 and 42 are the only combinations of 2 numbers whose sum is 6
nCr=7C2=21
2/21=probability that the sum of the numbers will be 6.
19 to 2 Odds in favor of the sum of the numbers on the two selected slips is not 6
.
Ed
RELATED QUESTIONS
Five slips of paper marked with the numbers 1, 2, 3, 4, and 5 are placed in a box and... (answered by solver91311)

Slips of paper marked with the numbers 1, 2, 3, 4 and 5 are placed in a box. After mixing (answered by stanbon)

The numbers 1 through 13 are written in separate slips of paper, and the slips are placed (answered by rothauserc)

The numbers 1 through 9 are written in separate slips of paper, and the slips are placed... (answered by ikleyn)

Six slips of paper with the numbers 6 through 11 are placed in a box. After mixing well,... (answered by ramkikk66)

Can someone please help me with this probability event. I know the solution but having... (answered by stanbon)

seven slips of paper with the numbers 7 through 13 are placed in a box. after mixng well, (answered by ewatrrr)

The numbers 1 through 10 are written in separate slips of paper, and the slips are... (answered by richard1234)

1)The numbers from, 1 to 8 are put on slips of paper and placed in a box. Three numbers (answered by stanbon)