A die is thrown twice. Find the probability that at least one of the two throws comes up with the number 5.

Hint: At least one throw with number 5 also means that both throws can have 5. Note all cases and all favourable outcomes in each case. Calculate their probabilities. Calculate the total probability as the sum of the probabilities in all the individual cases. This is the required answer.

Complete step by step answer:
A dice can have six outcomes that are (1,2,3,4,5,6).
For at least one throw with number 5 we have the following three cases.

1st case,
5 in 1st throw and(1,2,3,4,6) in second throw.
Probability= \[\dfrac{1}{6}\times \dfrac{5}{6}=\dfrac{5}{36}\]

2nd case,
(1,2,3,4,6) in 1st throw and 5 in 2nd throw.
Probability= \[\dfrac{5}{6}\times \dfrac{1}{6}=\dfrac{5}{36}\]

3rd case,
5 in 1st throw as well as 5 in 2nd throw.
Probability= \[\dfrac{1}{6}\times \dfrac{1}{6}=\dfrac{1}{36}\]

Total Probability= Probability in 1st case+ Probability in 2nd case+ Probability in 3rd case
\[=\dfrac{5}{36}+\dfrac{5}{36}+\dfrac{1}{36}=\dfrac{11}{36}\]

Note: For this type of question, students generally make mistakes in making cases for the outcomes. It can only be rectified by keeping all cases in mind.