There are equal numbers of even and odd numbers on each roll, so probability of SUM being odd or even is same so 1/2 each

Jack will need to roll the cube more than 2 times in order to get an even sum MEANS odd sum in first two rolls
1) first roll prob of odd = \(\frac{1}{2}\)
2) next roll prob of odd = \(\frac{1}{2}*\frac{1}{2}=\frac{1}{4}\)
so our answer = \(1-(\frac{1}{2}+\frac{1}{4})=1-\frac{3}{4}=\frac{1}{4}\)

B

Answer: B
Cube with 6 sides numbered 1 through 6.
More than 2 times to see sum of all of the rolls are even
P(1)= P(rolling once and not seeing even number) = 3/6 [if 1,3 or 5 happen]
P(2)= P(rolling twice and not seeing even number) = 9/36 = 1/4
first number is 1,3 or 5 and the second number should be in a way which sum of the first and the second is not even: [(1,2), (1,4),(1,6),(3,2), (3,4),(3,6),(5,1),(5,4),(5,6)
P(>2) = 1 - P(1) - P(2) = 1 - 1/2 - 1/4 = 1/4

A different way for the probability challenged folk like myself.

In order to have to roll more than twice, we have to look at the even/odd math properties. The question states that Jack stops rolling the dice when the sum is an even number, so what numbers added produce even numbers?

Odd + Odd = Even
Even + Even = Even

Odd + Even = Odd. Aha! This is what we are looking for!

From this property, we know now that the numbers rolled on the dice have to be opposite (If the first one is even, then the second roll has to be odd) and vice versa.

From here, We look at how many numbers can be even or odd on a dice, which is 3 out of 6, or 1/2.

1/2 x 1/2 = 1/4.

The answer is B.

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