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If you roll two fair six-sided dice, what is the probability that a least one dice shows a 3?​

 Mar 18, 2020
 #1
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If you roll two fair six-sided dice, what is the probability that a least one dice shows a 3?

 

\(\begin{array}{|r|r|r|r|r|r|r|r|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & & & \color{red}\text{x} \\ 2 & & & \color{red}\text{x} \\ 3 & \color{red}\text{x} & \color{red}\text{x} & \color{red}\text{x} & \color{red}\text{x} & \color{red}\text{x} & \color{red}\text{x} \\ 4 & & & \color{red}\text{x} \\ 5 & & & \color{red}\text{x} \\ 6 & & & \color{red}\text{x} \\ \hline \end{array} \)

 

The probability that a least one dice shows a 3 is  \(\mathbf{\dfrac{11}{36}}\quad (30.\overline{5}\ \%)\)

 

laugh

 Mar 18, 2020
 #2
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heureka, I understand what you did and it looks reasonable.  When Mr Barker explained how 1=2 in Algebra 1 that looked reasonable too.  I'm not saying your method is as specious as his was.  What I'm saying is, I've learned not to go by how something looks.  What I'm having a problem with is my inability to grasp why the probability wouldn't be simply twice the probability of rolling a 3 with one die.  Would you be willing to explain this for me?  I've been struggling to figure it out for a half hour and I'm no closer now than when I began.  Thank you for your help. 

Guest Mar 18, 2020
 #4
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I don't quite understand what your complaint is !
You simply figure out the probability of EXACTLY 1 three showing up as follows:
2C1 * 5^1 =2 * 5 =10
Then you would figure the probability of EXACTLY 2 threes showing up as follows:
2C2 * 5^0 =1 * 1 = 1
Then you would add the 2 probabilities together and that would give you "at least 1 three showing up", or: [10 + 1] / 6^2 =11 / 36 =30.556%

Guest Mar 18, 2020
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Thank you moderator for not printing my answer to this posting... NOT!  S***w you.

Guest Mar 19, 2020

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