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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?

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Hint: Let us first think about all possible outcomes when we throw a single dice and two dice together.
The outcomes when Peter throws the two dice together
(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) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
$\therefore $Total number of outcomes = 36
When Peter throws two dice, the favorable outcome for getting the product of numbers on the two dice equal to 25 is (5, 5).
Favorable number of outcomes = 1
$\therefore $Probability that Peter gets the product of numbers as 25 is
$$ \Rightarrow $$(Favorable number of outcomes)/ (Total number of outcomes) = (1/36)
The outcomes when Rina throws a dies are 1, 2, 3, 4, 5, and 6.
$\therefore $Total number of outcomes = 6
Rina throws a die and squares the number, so to get the number 25, the favorable outcome is 5.
Favorable number of outcomes = 1
$\therefore $Probability that Rina gets the square of the number as 25 is
(Favorable number of outcomes)/ (Total number of outcomes) = (1/6)
As, 1/6 >1/36, so Rina has a better chance to get the number 25 compared to Peter.

Note: A dice has numbers from 1 to 6. To get the product of two numbers as 25 on dice we only have one possibility that is 5, 5 on each dice. Probability is a measure of how often a particular event will happen. Probability of an event A is equal to (Number of favorable outcomes to A) / (Total Number of outcomes). Here, Peter throws two dice together that means he has a total 36 possible outcomes which is greater than the total number of possible outcomes for Rina, Which in turn decides the greater chance of getting the required condition.


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