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Jack rolls 5 fair six-sided dice. What is the probability that at least two dice show a 6?

 Dec 22, 2020
 #1
avatar+4 
0

6 are rolled. Probability of getting 6 is 1/6. We need at least 2. 1/6*1/6 but there are four left. We don't necessarily need 6 for these leftover 4 dice so that would by 5/6*5/6*5/6*5/6. Combining this with 1/6*1/6, that would be 1/6*1/6*5/6*5/6*5/6*5.6 which is equivalent to 0.01339591906.

 Dec 22, 2020
 #3
avatar+419 
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Make sure you include the at least part.

Pangolin14  Dec 22, 2020
 #2
avatar+419 
-1

Complementary counting

 

Case 1: No dice show 6

 

5^5/6^5

 

Case 2: One die shows a 6

 

1/6 * 5^4/6^4 = 5^4/6^5

 

Add them up to get 4026/6^5 = 671/1296

 Dec 22, 2020
 #4
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+1

[5 C 2 * 5^3] + [5 C 3 * 5^2] + [5 C 4 * 5^1] + [5 C 5 * 5^0]=1526 / 6^5==763 / 3888 ==19.62%. Or:

 

[5 C 2 * (1/6)^2 * (5/6)^3] + [5 C 3 * (1/6)^3 * (5/6)^2] + [5 C 4 * (1/6)^4 * (5/6)^1] + [5 C 5 + (1/6)^5 * (5/6)^0] =1526 / 6^5 =19.62%

 Dec 23, 2020
 #5
avatar+32013 
+3

Probability that no dice show a 6:  p1 = (5/6)^5

Probability that exactly one die shows a six:  p2 = 5*(1/6)*(5/6)^4

Probability that at least two dice show a six: p = 1 - p1 - p2 ≈ 0.196

 Dec 23, 2020

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