two six - sided dice are tossed. event a: the first die does not land on 5. event b: the second die lands on…

two six - sided dice are tossed. event a: the first die does not land on 5. event b: the second die lands on 4. what is the probability that both events will occur? for independent events: p(a and b)=p(a)·p(b) p(a and b)=? give your answer in simplest form
Answer
Answer:
$\frac{5}{36}$
Explanation:
Step1: Calculate P(A)
The first die has 6 possible outcomes. The event A is that the first die does not land on 5. So there are 5 favorable outcomes. Then $P(A)=\frac{5}{6}$.
Step2: Calculate P(B)
The second die has 6 possible outcomes. The event B is that the second die lands on 4. So there is 1 favorable outcome. Then $P(B)=\frac{1}{6}$.
Step3: Calculate P(A and B)
Since A and B are independent events, $P(A\text{ and }B)=P(A)\cdot P(B)$. Substitute $P(A)=\frac{5}{6}$ and $P(B)=\frac{1}{6}$ into the formula, we get $P(A\text{ and }B)=\frac{5}{6}\times\frac{1}{6}=\frac{5}{36}$.