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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  [#permalink]

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New post Updated on: 03 Mar 2015, 05:31
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E

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  85% (hard)

Question Stats:

49% (01:59) correct 51% (02:09) wrong based on 130 sessions

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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

Originally posted by amianik on 02 Mar 2015, 13:15.
Last edited by Bunuel on 03 Mar 2015, 05:31, edited 1 time in total.
Edited the question.
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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube  [#permalink]

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New post 02 Mar 2015, 14:55
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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:

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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube  [#permalink]

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New post 02 Mar 2015, 21:55
EMPOWERgmatRichC wrote:
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:

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


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
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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube  [#permalink]

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New post 03 Mar 2015, 00:36
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.

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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube  [#permalink]

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New post 03 Mar 2015, 05:32
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


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
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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube  [#permalink]

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New post 16 Nov 2017, 05:27
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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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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube  [#permalink]

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New post 30 Nov 2019, 02:12
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



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Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube  [#permalink]

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New post 01 Dec 2019, 14:17
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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


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

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Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube   [#permalink] 01 Dec 2019, 14:17
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