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

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Jack has a cube with 6 sides numbered 1 through 6. He rolls [#permalink] New post   18 Feb 2020, 16:39
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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
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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls [#permalink] New post   22 Feb 2020, 15:33
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Post already discussed at this link https://gre.myprepclub.com/forum/jack-has- ... 11273.html

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Jack has a cube with 6 sides numbered 1 through 6. He rolls [#permalink] New post   22 Mar 2021, 06:49
solution plz....
according to me , i took 0.50*o.5*0.5=1/8 but ans is 0.5*0.5 why
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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls [#permalink] New post   22 Mar 2021, 22:40
void wrote:
solution plz....
according to me , i took 0.50*o.5*0.5=1/8 but ans is 0.5*0.5 why


according to me , i took 0.50*o.5*0.5=1/8 but ans is 0.5*0.5 why

The question is saying more than two i.e. 3, 4, 5 ..... , what you are doing is exactly 3.

The simplest way to solve this question is:

P (more than 2) = 1 - P (only 1) - P (only 2)

Lets find P (only 1) = \(\frac{3}{6} = \frac{1}{2}\)

and, P (only 2) = \((\frac{1}{2})(\frac{1}{2}) = \frac{1}{4}\)

Therefore, P (more than 2) = \(1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\)
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Re: Jack has a cube with 6 sides numbered 1 through 6. He rolls [#permalink]
22 Mar 2021, 22:40
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