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

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

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New post 24 Nov 2019, 10:45
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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) \(12(1/6)^2(5/6)^2 + 6(1/6)^3(5/6) + (1/6)^4\)
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The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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New post 13 Dec 2019, 21:10
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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
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Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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New post 01 Dec 2019, 02:30
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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 \(6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4\)
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Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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New post 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
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The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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New post 13 Dec 2019, 21:33
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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)
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Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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New post 13 Dec 2019, 22:53
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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 = \(6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4\)=\(\frac{6*1^2*5^2+4*1^3*5+1^4}{6^4}=\frac{150+20+1}{6^4}=\frac{171}{6^4}\)
Thus our answer
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Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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