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Jack has a cube with 6 sides numbered 1 through
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Updated on: 14 Oct 2018, 03:56
Updated on: 14 Oct 2018, 03:56
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Question Stats:
53% (01:44) correct
46% (01:34) wrong based on 139 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
(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 3/4
Re: Please help me to solve this Probability question.
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13 Oct 2018, 11:37
13 Oct 2018, 11:37
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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
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.
General Discussion
Re: Jack has a cube with 6 sides numbered 1 through
[#permalink]
18 Oct 2018, 05:50
18 Oct 2018, 05:50
1
Expert Reply
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
(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 3/4
There are equal numbers of even and odd numbers on each roll, so probability of SUM being odd or even is same so 1/2 each
Jack will need to roll the cube more than 2 times in order to get an even sum MEANS odd sum in first two rolls
1) first roll prob of odd = \(\frac{1}{2}\)
2) next roll prob of odd = \(\frac{1}{2}*\frac{1}{2}=\frac{1}{4}\)
so our answer = \(1-(\frac{1}{2}+\frac{1}{4})=1-\frac{3}{4}=\frac{1}{4}\)
B
