A dice is rolled twice. What is the probability of observing at least one five?

Melody is correct......

Here is another way to look at it

Probability of NOT rolling a 5 on the first roll is 5/6

Probability of NOT rolling a 5 on the second roll is also 5/6

Probability of NOT rolling a 5 in two rolls is then   5/6 x 5/6 = 25 /36

so the probability of  (YES) rolling a 5 is   36/36 - 25/36 = 11/36  with two rolls....

 One of these days I will understand probability.....'probably'? or Maybe ? LOL

(I think !)

Each time you roll the die (one is a 'die'  ...two are called 'dice') you have a 1/6 chance of rolling a '5'

rolling TWO times DOUBLES that chance    2 x 1/6 = 1/3 chance of rolling a '5'

A dice is rolled twice. What is the probability of observing at least one five?

P(first one 5 )+P(both 5) - P(both 5)

I'm going to try doing it a couple of different ways and see I get the same answers :)

\(\frac{1}{6}+\frac{1}{6}-\frac{1}{6}*\frac{1}{6}\\=\frac{12}{66}-\frac{1}{36}\\=\frac{11}{36}\\~\\ or\\ \frac{1}{6}*\frac{5}{6}+\frac{5}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}\\ =\frac{5}{36}+\frac{5}{36}+\frac{1}{6}\\=\frac{11}{36}\)

A dice is rolled twice. What is the probability of observing at least one five?

\(\color{green}{\text{In green dice first rolling }}\\ \color{blue}{\text{In blue dice second rolling }}\\ \color{red}{\text{In red number } 5 } \)

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

\(\text{The probability of rolling at least one five is } \dfrac{11}{36} \text{ or } 30.\bar{5}\ \%\)