Jack has a cube with 6 sides numbered 1 through 6

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

Jack will not get the sum even in the first two rolls if in the first roll, he gets an odd number and in the second roll, he gets an even number.

So, the probability of getting an odd number = 3/6 = 1/2; note that there are three odd numbers (1, 3, 5) out of 1, 2, 3, 4, 5, and 6.

Similarly, the probability of getting an even number = 3/6 = 1/2; note that there are three even numbers (2, 4, 6) out of 1, 2, 3, 4, 5, and 6.

So, the probability that Jack will not an even sum in the first two rolls = 1/2*1/2 = 1/4

The correct answer: B

Hope this helps!

-Jay
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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

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

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