Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube

Question

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

(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 3/4

shashankism · Accepted Answer

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

BrentGMATPrepNow · Answer

Let's apply the  complement  property

P(it takes Jack MORE THAN 2 rolls to get even sum) = 1 -  P(it takes 2 rolls or fewer to get even sum)

P(it takes 2 rolls or fewer to get even sum)  
There are exactly two ways in which it can take Jack 2 rolls or fewer to get an even sum:
- Jack rolls an even number on the 1st roll
- Jack rolls an odd number on the 1st roll and then an odd number on the 2nd roll

So, P(it take 2 rolls or fewer to get even sum) = P(even on 1st roll   OR    odd on first AND odd on 2nd)
= P(even on 1st roll)   +    P(odd on first AND odd on 2nd)
= 1/2   +    (1/2)(1/2)
= 1/2   +    1/4
=  3/4

So, P(it takes Jack MORE THAN 2 rolls to get even sum) = 1 -  3/4 
= 1/4

Answer: B

Cheers
Brent

EMPOWERgmatRichC · Answer

Hi amianik,

This question asks us for the probability of a specific result: needing MORE than 2 rolls to get a sum that is EVEN. As a probability question, we'll need to multiply the individual results of each "step" to figure out the overall probability.

Let's work through the "events" that would need to occur...

IF....
The first roll is EVEN, then we're done.
To end up with MORE than 2 rolls though, the first roll would have to be ODD.
3/6 outcomes on the first roll are odd. 3/6 = 1/2

Next, if the second roll is ODD, then we're done (since Odd + Odd = Even).
To end up with MORE than 2 rolls, the second roll would have to be EVEN
3/6 outcomes on the second roll are even. 3/6 = 1/2

At this point, if the first roll is odd and the second roll is even, then we would NEED another roll (meaning MORE than 2 rolls)...
(1/2)(1/2) = 1/4 of the possible outcomes on the first 2 rolls would require at least one additional roll.

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

amianik · Answer

but if the 3rd roll is even then the sum will be odd, so he needs to roll fourth time.. but here the ans has been demonstrated as probability of even and odd

EMPOWERgmatRichC · Answer

Hi amianik,

Since the question asks for the probability of rolling MORE than 2 rolls, we don't have to think beyond the second roll. As long as we know that it's MORE than 2 rolls.....3 rolls, 4 rolls, 5 rolls etc.....then they're all part of the same group. By rolling an odd first, then an even second, we know that there will be at least one more roll. ALL of those options fall into what we're asked for - and THAT total is 1/4 of the possibilities.

GMAT assassins aren't born, they're made,
Rich

Bunuel · Answer

Similar (though harder) questions to practice: molly-is-rolling-a-number-cube-with-faces-numbered-1-to-162116.html a-cube-with-its-sides-numbered-1-through-6-is-rolled-twice-131460.html

hudacse6 · Answer

GMATPrepNow 
Better if you have a look.

ScottTargetTestPrep · Answer

We can use the fact that:

P(even sum in more than 2 rolls) = 1 - P(even sum in 1 roll) - P(even sum in 2 rolls)

Since P(even sum in 1 roll) = 1/2, and

P(even sum in 2 rolls) = P(odd sum in 1st roll and odd sum in 2nd roll) = 1/2 x 1/2 = 1/4,

so P(even sum in more than 2 rolls) = 1 - 1/2 - 1/4 = 1/4.

Answer: B

bumpbot · Answer

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Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube

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Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube

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