Hi! I got stuck on this hard probability problem while doing my math homework. I would appreciate any help!
A circular spinner is divided into 4 congruent sectors, each marked with a positive integer from 1 to 4. If the spinner is spun 3 times, what is the probability that the product of all the numbers spun is an even number? Express your answer as a common fraction.
I have tried creating different scenarios such as even × odd × odd, odd × even × odd, etc. However, I couldn't reach an answer :(
Have a great day!
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
which is 1/2*1/2*1/2=1/8 chance of odd, so 7/8 chance of even