A dice is thrown twice. The probability of ge | vedclass

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(B) Let $A$ be the event of getting $4, 5,$ or $6$ in the first throw. The number of favorable outcomes is $3$. Total outcomes = $6$.$P(A) = \frac{3}{

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$\frac{1}{3}$

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(B) Let $A$ be the event of getting $4, 5,$ or $6$ in the first throw. The number of favorable outcomes is $3$. Total outcomes = $6$.$P(A) = \frac{3}{6} = \frac{1}{2}$.Let $B$ be the event of getting $1, 2, 3,$ or $4$ in the second throw. The number of favorable outcomes is $4$. Total outcomes = $6$.$P(B) = \frac{4}{6} = \frac{2}{3}$.Since the two throws are independent events,the probability of both occurring is $P(A \cap B) = P(A) 	imes P(B)$.$P(A \cap B) = \frac{1}{2} 	imes \frac{2}{3} = \frac{1}{3}$.

$A$ dice is thrown twice. The probability of getting $4, 5$ or $6$ in the first throw and $1, 2, 3$ or $4$ in the second throw is

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