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Meyer rolls two fair, ordinary dice with the numbers  on their sides. What is the probability that at least one of the dice shows a square number?

 Jan 16, 2019
 #1
avatar+18337 
0

Is 4 the only thing you consider a square number?          Or is six kinda square?    Or is '1' a square (1^2 = 1...so one is a square)...

   need clarification   .....do you think 5 is a square on the die?

 Jan 16, 2019
 #2
avatar+5172 
+3

\(\text{The only two numbers appearing on each die that are perfect squares are }1 \text{ and } 4\\ P[\text{1 or 4 appear on at least 1 die out of 2}] = 1 - P[\text{no 1s or 4s}] = \\ 1-\dfrac{4^2}{36} = \dfrac{20}{36} = \dfrac{5}{9}\)

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 Jan 16, 2019
 #3
avatar+22363 
+8

Meyer rolls two fair, ordinary dice with the numbers  on their sides.
What is the probability that at least one of the dice shows a square number?

 

\(\color{red}\text{square number} \)

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

 

\(\text{The probability is $\dfrac{20}{36} = \dfrac{5}{9} \quad (55.6\ \%) $ } \)

 

laugh

 Jan 16, 2019

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