If a dice is thrown twice,then the probabilit | vedclass

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(C) When a dice is thrown,the total number of possible outcomes is $6$. The probability of getting $1$ in a single throw is $\frac{1}{6}$.The probabil

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$\frac{5}{36}$

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(C) When a dice is thrown,the total number of possible outcomes is $6$. The probability of getting $1$ in a single throw is $\frac{1}{6}$.The probability of $NOT$ getting $1$ in a single throw is $1 - \frac{1}{6} = \frac{5}{6}$.Since the two throws are independent events,the probability of getting $1$ in the first throw $AND$ not getting $1$ in the second throw is calculated as follows:Required Probability $= P(	ext{getting } 1 	ext{ in first throw}) 	imes P(	ext{not getting } 1 	ext{ in second throw})$Required Probability $= \frac{1}{6} 	imes \frac{5}{6} = \frac{5}{36}$.

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

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