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(C) The total number of outcomes when a dice is thrown twice is $6 	imes 6 = 36$.Let $E_1$ be the event of getting $1$ in the first throw. The outcom

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$\frac{5}{36}$

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(C) The total number of outcomes when a dice is thrown twice is $6 	imes 6 = 36$.Let $E_1$ be the event of getting $1$ in the first throw. The outcomes are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$.Let $E_2$ be the event of not getting $1$ in the second throw. The outcomes for the second throw can be $2, 3, 4, 5, 6$.We want the probability of getting $1$ in the first throw $AND$ not getting $1$ in the second throw.The favorable outcomes are $(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$,which are $5$ outcomes.The required probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{5}{36}$.

If a dice is thrown twice,then the probability of getting $1$ in the first throw only is

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