Problem Teachers' Notes Hint Solution

# Odds and Evens

Congratulations to Sam, James and Esther who sent in correct solutions to this problem.

Several others of you were really close, but made slight errors.

Here is Esther's solution:

Set D would give the greatest proportion of even numbers.
The possibilities for this set are:
(1,3) (1,4) (1,5) (1,7) (1,9)
(3,4) (3,5) (3,7) (3,9)
(4,5) (4,7) (4,9)
(5,7) (5,9)
(7,9)
of which 10 out of 15 are even, a probability of 2/3 .
When you work out A in this way, the probability is 2/5, B is 3/5 and C is 7/15.

Esther correctly spotted that if you could select any numbers with at least one odd, the best numbers would be if they were all odd - two odds always add to make an even .

If assume that picking 3 and 5 is same as picking 5 and 3, then: A - 4 out of possible 10 are even: 3+5, 2+6, 2+4, 6+4 B - 6 out of possible 10 are even: 3+1, 3+5, 3+7, 1+5, 1+7, 5+7 C - 7 out of possible 15 are even: 8+2, 8+6, 8+4, 3+5, 2+6, 2+4, 4+6 D - 10 out of possible 15 are even: 7+9, 7+3, 7+1, 7+5, 3+9, 3+1, 3+5, 9+1, 9+5, 5+1 If all five balls or all six balls were odd, every combination would give an even total.