When a die is thrown twice,what is the probab | vedclass

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(B) Let $A$ be the event of getting $4, 5,$ or $6$ on the first throw. The number of favorable outcomes is $3$. The total outcomes for one die is $6$.

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

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

When a die is thrown twice,what is the probability of getting $4, 5,$ or $6$ on the first throw and $1, 2, 3,$ or $4$ on the second throw?

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