BTGModeratorVI wrote: ↑Wed May 13, 2020 10:57 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
