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?

Guest Jan 16, 2019

#1**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?

ElectricPavlov Jan 16, 2019

#2**+3 **

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

.Rom Jan 16, 2019

#3**+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?**

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

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

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

heureka Jan 16, 2019