amianik 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

We need to find the probability that Jack will need to roll the cube more than 2 times to get even sum..

For this we will find the probability that Jack will need to roll the cube 2 or less than 2 times to get even sum..

Case 1 : He gets even no. in 1st roll

P1 = 3/6 = 1/2 (Rolls may be 2,4,6)

Case 2 : (He gets odd in 1st roll and another odd in 2nd roll to make the sum even.)

P2 = 3*3/6*6 (e.g. 1,1;1,3;1,5;......;5,5) = 1/4

Total probability that Jack will need to roll the cube 2 or less than 2 times to get even sum = 1/2 + 1/4 = 3/4

Probability that Jack will need to roll the cube more than 2 times to get even sum. = 1-3/4 = 1/4

Answer B
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