It was easier to me to figure out what the probability first of actually getting two consecutive numbers;
First the odds on any given roll of getting an even number is 3/6 or 1/2. The same is true of odd numbers. Because there is a 1/2 for even roll of getting an even, and a 1/2 change of getting and odd, we can think of this largely as a coin game. Now I must figure out how many possible combinations of even and odd there are in 4 rolls.
There are 5 cases; o=odd, e=even
1) All odd; oooo, the number of possible combinations = 4!/4! = 1
2)1 even the rest odd; eooo, the number of combinations with this is = 4!/(1!)(3!)= 4
3) 2 odd 2 even; ooee, number of combinations = 4!/(2!*2!) = 3*2*1 = 6
4)3 even 1 odd; eeeo, number of combinations = 4!/(1!3!)4= 4
5) All even; eeee, number of combinations = 4!/4! = 1
So, total there are 1+4+6+4+1=16 combinations. This could also be determine this by using the counting principle, which says that if you trying count the number of possible combinations, simply multiple the number of possiblities for each decision together. With this problem each roll will result in an even or odd number, so roll has 2 possibilities. So if there are 4 rolls;
2*2*2*2=16 (for four rolls)
Beacuse there are only 4 rolls and 2 possible results for each roll, it's easy to determine combinations where you would have 2 consecutive evens.
1) eeoo
2) oeeo
3) ooee
4) eeeo
5) oeee
6) eeee
These are the 6 ways in which you can roll consecutive even numbers. So there are 10 ways in which you can roll the dice and not get 2 consecutive even numbers.
10/16=.625, but this is not an answer choice.
There something missing from the question, or is my math above not correct?
Thanks,
Jared
