1 is odd. An even number can be divided evenly, even though some elementary math teachers consider it even.
No. The sum of 1 odd number and 1 even number must always be odd.
You want: P(even) * P(even) * P(even) * P(even). P(even) = 3/6 or 1/2. Therefore probability even on 4 rolls is: 1/2 * 1/2 * 1/2 * 1/2 = 1/16 = 0.0625.
When a die is cast half the possible numbers are even. Therefore the probability is:Probability that die 1 is even = 1/2Probability that die 2 is even = 1/2Probability that both are even = 1/2 * 1/2 = 1/4.
1 in 4. Your choices are to have an odd or even total, which is a 1 in 2 chance. In order to get an even number, you have to have both die even or odd. So that cuts the 1 in 2 in half, giving you a 1 in 4 chance of both dice being even. (1 in 2 for even times 1 in 2 both even gives you 1 in 4.)
Any odd number is of the form 2n + 1. Adding two of these, call them a and b, will give even numbers (a-1) + 1 and (b-1) + 1. Since you then have 2 even numbers plus (1 +1), you have another even nunmber.
-- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number plus an extra ' 1 '. -- Add them. You get (an even number) + (an even number) + ( 2 ). That's an even number. ======================== Another mind-bending , brain-busting way to look at it : -- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number minus ' 1 '. -- Add them. You get (an even number) + (an even number) + ( [1 - 1] or zero ). That's an even number.
yes, the product of 2 odd numbers is always an odd number. Well, the question is why. The first number is "even" + 1. Multiply both of these by your odd number. Now the "even" times "odd" is even, because every "1" in the odd number becomes a "2". And then the remaining 1 times "odd" must be odd, which is an even +1. Add it all up and you get evens everywhere except that final "1". So the result is even + 1 which is odd. There is a quicker way if you know how to multiply bracketed terms: odd x odd = (even + 1)x(even +1)= even x even +even +even +1 = must be odd.
20 is even so all multiples of 20 must be even. 1 is not even.
The two dice can be (odd, odd), (odd, even), (even, odd), or (even, even). Thus the probability of two dice totaling an even number is 2*(1/2)*(1/2) = 1/2. 50/50
odd x even = even (2n + 1)(2n) = 2[n(2n + 1)] let n(2n + 1) = t = 2t (even)
*1 is an odd number * 0 is an even number