GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 20 Jan 2020, 20:29

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The probability of getting a '6' on a role of die is 1/6. If this die

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Status: So far only Dreams i have!!
Joined: 05 Jan 2015
Posts: 216
WE: Consulting (Consulting)
The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

### Show Tags

24 Nov 2019, 10:45
1
00:00

Difficulty:

45% (medium)

Question Stats:

67% (02:05) correct 33% (02:14) wrong based on 52 sessions

### HideShow timer Statistics

The probability of getting a '6' on a role of die is 1/6. If this die is rolled
4 times, what is the probability of getting a '6' at least 2 times?

(A) $$(1/6)^4$$
(B) $$2(1 /6)^3(5/6) + (1/6)^4$$
(C) $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(D) $$12(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(E) $$12(1/6)^2(5/6)^2 + 6(1/6)^3(5/6) + (1/6)^4$$
Math Expert
Joined: 02 Aug 2009
Posts: 8344
The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

### Show Tags

13 Dec 2019, 21:10
1
1
fauji wrote:
The probability of getting a '6' on a role of die is 1/6. If this die is rolled
4 times, what is the probability of getting a '6' at least 2 times?

(A) $$(1/6)^4$$
(B) $$2(1 /6)^3(5/6) + (1/6)^4$$
(C) $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(D) $$12(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(E) $$12(1/6)^2(5/6)^2 + 6(1/6)^3(5/6) + (1/6)^4$$

Ok let us see the solution in a bit detail

The probability of getting a particular number is $$\frac{1}{6}$$ and NOT getting is $$\frac{5}{6}$$.

1. 2 times
So 2 times the probability will be $$\frac{1}{6}$$ and other two times $$\frac{5}{6}$$ as any of the 5 can be chosen
Now this 2 times can be chosen in 4C2 ways, so $$4C2*(\frac{1}{6})^2*(\frac{5}{6})^2=6*(\frac{1}{6})^2*(\frac{5}{6})^2$$
2. 3 times
So 3 times the probability will be $$\frac{1}{6}$$ and one times $$\frac{5}{6}$$ as any of the 5 can be chosen
Now this 3 times can be chosen in 4C3 ways, so $$4C3*(\frac{1}{6})^3*(\frac{5}{6})^1=4*(\frac{1}{6})^3*(\frac{5}{6})$$
3. 4 times
So all 4 times the probability will be $$\frac{1}{6}$$.
Now this 4 times can be chosen in 4C4 ways, so $$4C4*(\frac{1}{6})^4=(\frac{1}{6})^4$$
Add all 3 and you will get the answer

C
_________________
##### General Discussion
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5700
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

### Show Tags

01 Dec 2019, 02:30
1
fauji wrote:
The probability of getting a '6' on a role of die is 1/6. If this die is rolled
4 times, what is the probability of getting a '6' at least 2 times?

(A) $$(1/6)^4$$
(B) $$2(1 /6)^3(5/6) + (1/6)^4$$
(C) $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(D) $$12(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(E) $$12(1/6)^2(5/6)^2 + 6(1/6)^3(5/6) + (1/6)^4$$

total ways ; 4c2 * ( 5/6)^2*(1/6)^2 + 4c1* (1/6)^3*(5/6) + (1/6)6
IMO C $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
Intern
Joined: 22 Oct 2018
Posts: 18
Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

### Show Tags

13 Dec 2019, 20:08
Could someone explain the solution to this problem ?

I tried doing it using the 1 - P(6 occurring only once + Not at all) and I am getting a really huge and lengthy answer
Rice (Jones) School Moderator
Joined: 18 Jun 2018
Posts: 329
Location: United States (AZ)
Concentration: Finance, Healthcare
GMAT 1: 600 Q44 V28
GPA: 3.36
The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

### Show Tags

13 Dec 2019, 21:33
2
SiddharthR Here are the possible options: 66NN, 666N, and 6666. Where N = not 6
We are given P(6) = $$\frac{1}{6}$$ ==> P(N) = $$\frac{5}{6}$$
Therefore P(at least two 6s) = 66NN + 666N + 6666
Note for 66NN we have $$\frac{4!}{2!2!}$$ ways = 6 ways to arrange that option , and for 666N, we have $$\frac{4!}{3!}$$ ways = 4 ways
Therefore, P(at least 2 6s) = 6($$\frac{1}{6}$$)^$${2}$$($$\frac{5}{6}$$)^$${2}$$ + 4($$\frac{1}{6}$$)^$${3}$$($$\frac{5}{6}$$) + ($$\frac{1}{6}$$)^$${4}$$ (the answer is C)
Math Expert
Joined: 02 Aug 2009
Posts: 8344
Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

### Show Tags

13 Dec 2019, 22:53
1
SiddharthR wrote:
Could someone explain the solution to this problem ?

I tried doing it using the 1 - P(6 occurring only once + Not at all) and I am getting a really huge and lengthy answer

SiddharthR if you want to do with your formula, then you have to find the final solution and similarly look for that answer in the choices after simplifying them..

1) Occurring once - $$4C1*(\frac{1}{6})^1*(\frac{5}{6})^3=\frac{4*1*5^3}{6^4}=\frac{500}{6^4}$$ as explained in my solution above
2) Not at all - $$4C0*(\frac{1}{6})^0*(\frac{5}{6})^4=\frac{4*1*5^4}{6^4}=\frac{625}{6^4}$$
Total = $$\frac{500}{6^4}+\frac{625}{6^4}=\frac{1125}{6^4}$$
Our answer= $$1-\frac{1125}{6^4}=\frac{1296-1125}{6^4}=\frac{171}{6^4}$$

Now, choice C = $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$=$$\frac{6*1^2*5^2+4*1^3*5+1^4}{6^4}=\frac{150+20+1}{6^4}=\frac{171}{6^4}$$
_________________
Intern
Joined: 22 Oct 2018
Posts: 18
Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

### Show Tags

15 Dec 2019, 15:13
I get it now. Thank you for explaining it to me
Re: The probability of getting a '6' on a role of die is 1/6. If this die   [#permalink] 15 Dec 2019, 15:13
Display posts from previous: Sort by

# The probability of getting a '6' on a role of die is 1/6. If this die

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne