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

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BTGModeratorVI: Legendary Member; Posts: 1223; Joined: Sat Feb 15, 2020 2:23 pm; Followed by:1 members

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

by BTGModeratorVI » Thu Jul 23, 2020 6:48 am

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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

Answer: B
Source: Manhattan prep

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Source: Beat The GMAT — Problem Solving | Thu Jul 23, 2020 6:48 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

Re: Jack has a cube with 6 sides numbered 1 through 6.

by Brent@GMATPrepNow » Sat Jul 25, 2020 6:39 am

BTGModeratorVI wrote: ↑
Thu Jul 23, 2020 6:48 am
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

Answer: B
Source: Manhattan prep

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

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

Re: Jack has a cube with 6 sides numbered 1 through 6.

by Scott@TargetTestPrep » Sun Jul 26, 2020 8:47 am

BTGModeratorVI wrote: ↑
Thu Jul 23, 2020 6:48 am
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

Answer: B

Solution:

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

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