Question

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 that the sum of the numbers on the two slips is 8?

Answer 1

just did this one last night i think

Answer 2

there are \(\dbinom{7}{2}=\frac{7\times 6}{2}=21\) ways to choose two numbers out of those seven (order not counting)
of those 21 ways, only 2 give a total of 8, and 19 do not

Answer 3

ok that is wrong
3 give a total of 8, 18 do not

Answer 4

answers include:
1 to 7
B

1 to 9
C

1 to 6
D

2 to 21

so would it be 2 to 21?

Answer 5

1+7,2+6,3+5 give 8 i miscounted
in any case three do, eighteen do not, odds in favor are \(3:18\) or maybe if you reduce you get \(1:6\)

Answer 6

nevermind!!!

Answer 7

okay cool thank you!!

Answer 8

yw

Answer 9

the next question says 5 slips of paper and goes up to numbers 1, 2, 3, 4, and 5. if this is the case, then would it be 5/2 as well? I'm confused to how you got (7/2) = 7 *6 /2

Answer 10

i was not computing \(\frac{7}{2}\) but rather \(\binom{7}{2}\)
you may have seen this written as \(_7C_2\)

Answer 11

ohh!! okay that makes a lot more sense

Answer 12

in this case you are choosing 2 numbers out of 5
number of ways to do this is \(_5C_2=\frac{5\times 4}{2}=10\)

Answer 13

what is the sum supposed to be?

Answer 14

still 8 or some other number?