For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If

This topic has expert replies
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pm
Followed by:23 members

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4


OA D

Source: Manhattan Prep
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8083
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members
BTGmoderatorDC wrote:
Tue May 12, 2020 9:26 pm
For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4


OA D
“At least 3 twos in 4 rolls” means we could have 3 twos and 1 non-two number OR all 4 twos.

P(3 twos and 1 non-two number) = 4C3 x (1/6)^3 x (5/6) = 4(1/6)^3 (5/6)

P(all 4 twos) = 4C4 x (1/6)^4 = (1/6)^4

Therefore, P(at least 3 twos) = 4(1/6)^3 (5/6) + (1/6)^4

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage