Hard Probability Problem

Rewritten cause I got in a mess.

It will be even if one or more of the numbers is even.  (2 will be a factor)

So it will only be odd if all thre are odd  and the cances of that happening is 1/8

Prob of even = 1 -1/8  = 7/8

Just like mathsProblemSolver has already said.  Thanks MPS

Hi! The answer would be a fraction; the numerator being the number of specific possibilies (product of numbers spun being even), and the denomerator being the total possibilities. Since there are 4 sectors and the spinner is spun 3 times, the denomerator would be 4 * 4 * 4 = 64. Using what you previously said about the different scenarios with even and odd numbers, you quickly find out that the only way for the product of 3 numbers to be odd is when all three numbers are odd (leaving the rest of the probabilitys [even x even x even, even x even x odd, and odd x odd x even] to result in even numbers). Since it is easier to find the probability of the product being odd, we can do that and then subtract our answer from the total at the end. In all the spins, 2 numbers out of the 4 are odd, so there are 2 options per spin. That means that there are 2 * 2 * 2 = 8 possibilities. (1*1*1, 1*1*3, 1*3*1, 1*3*3, 3*1*1, 3*1*3, 3*3*1, and 3*3*3). That means that 64-8=56 of the 64 spins will be even, so the answer is 56/64 or 7/8....I think at least

Instead of casework, try complementary.

P(odd) happens when it is odd * odd = (1/2)^3 = 1/8 (all odd)

1-1/8 = 7/8

odd*odd*odd =odd 

which is 1/2*1/2*1/2=1/8 chance of odd, so 7/8 chance of even

Thanks for all of your answers/suggestions! It helped me greatly!