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(B) Apoorv throws two dice once.Total number of outcomes $= 6 	imes 6 = 36$.The only way to get a product of $36$ is $(6, 6)$.Number of favorable out

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Peehu

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(B) Apoorv throws two dice once.Total number of outcomes $= 6 	imes 6 = 36$.The only way to get a product of $36$ is $(6, 6)$.Number of favorable outcomes $= 1$.Probability for Apoorv $= \frac{1}{36}$.Peehu throws one die.Total number of outcomes $= 6$.The only way to get a square of $36$ is by rolling a $6$ $(6^2 = 36)$.Number of favorable outcomes $= 1$.Probability for Peehu $= \frac{1}{6} = \frac{6}{36}$.Comparing the probabilities,$\frac{6}{36} > \frac{1}{36}$.Therefore,Peehu has a better chance of getting the number $36$.

Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number $36$? Why?

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