Sawant91 wrote:
Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube repeatedly until the first time that the sum of
all of his rolls is even, at which time he stops. (Note: it is possible to roll the cube just once.) What is the
probability that Jack will need to roll the cube more than 2 times in order to get an even sum?
(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 3/4
You can say that the probability that Jack rolls more than twice is 1- Probability Jack rolls once or twice.
Probability Jack rolls once = \(P(1)= \frac{3}{6}=\frac{1}{2}\)
Here is the possible solution of 2 roll even numbers {1,1},{1,3},{1,5},{3,1},{3,3},{3,5}.........{5,5} or 9 combinations out of 36
Probability Jack rolls twice = \(P(2)= \frac{9}{36}=\frac{1}{4}\)
Probability Jack rolls once or twice= Probability Jack rolls once + Probability Jack rolls twice=\(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
Hence the probability that Jack rolls more than twice = 1- Probability Jack rolls once or twice \(= 1- \frac{3}{4}=\frac{1}{4}\).
Hence B is the answer.