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

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