Question

Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number $25$?

Answer 1

Let us first write the all possible oucomes when Peter throws two different dice together.

$\begin{array}{c}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(5,3),(5,4),(5,5),(5,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}$

$\begin{array}{c}\therefore \text{Total number of outcomes}=36\\ \text{The favorable outcome for getting the product of numbers on the dice equalto}25\text{is}(5,5)\text{.}\\ \text{Favourable number of outcomes}=1\end{array}$

$\begin{array}{c}\therefore \text{Probability that Peter gets the product of numbers as}25\\ =\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}=\frac{1}{36}\\ \text{The outcomes when Rina throws a die are}1,2,3,4,5,6\end{array}$

$\begin{array}{c}\therefore \text{Total number of outcomes}=6\\ \text{Rina throws a die and squares the number, so to get the number}25,\text{the favourable outcome is}\\ \text{5.}\\ \text{Favourable number of outcomes =}1\end{array}$

$\begin{array}{c}\text{Favourable number of outcomes = 1}\\ \therefore \text{Probability that Rina gets the square of the number as}25\\ \frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}=\frac{1}{6}\\ \text{As,}1/6>1/36,\text{so Rina has better chance to get the number}25\text{.}\end{array}$