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

Question

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

chetan2u · Accepted Answer

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

Archit3110 · Answer

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

SiddharthR · Answer

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

bebs · Answer

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  )

chetan2u · 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} 
Thus our answer

SiddharthR · Answer

I get it now. Thank you for explaining it to me

luisdicampo · Answer

Deconstructing the Question

Topic: Binomial Probability

We need the probability of getting a '6'  at least 2 times  in 4 rolls.
This is the sum of three mutually exclusive cases:
 P(At least 2) = P(Exactly 2) + P(Exactly 3) + P(Exactly 4)

We can identify the correct answer simply by calculating the  Combinations (coefficients)  for each term.
Formula:  nCk * (success)^k * (failure)^(n-k)

1. P(Exactly 2)  
We choose 2 rolls out of 4 to be a '6'.
Coefficient =  4C2 = (4*3)/2 =    6 
Term:  6 * (1/6)^2 * (5/6)^2

2. P(Exactly 3)  
We choose 3 rolls out of 4.
Coefficient =  4C3 =    4 
Term:  4 * (1/6)^3 * (5/6)^1

3. P(Exactly 4)  
We choose all 4 rolls.
Coefficient =  4C4 =    1 
Term:  1 * (1/6)^4

4. Match with Options  
We are looking for the sum where the coefficients are  6, 4, and 1 .
Option  (C)  matches this pattern perfectly:
 6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4

The correct answer is C.

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