We have two picking rounds with replacement.
We have to calculate the probability of picking a 7 and then picking an even number.
As the picking is made with replacement, the probabilities of both events are independent.
Then, we can write:
![P(7\&E)=P(7)\cdot P(E)](https://img.qammunity.org/2023/formulas/mathematics/college/lb5uum5l2j5q20uqk7wvdf9f6bmn46sszj.png)
We can calculate the probability of each event as the quotient between the successful outcomes and the possible outcomes.
For the first event we have 1/4 as we have one successful event (getting a 7) out of 4 possible events.
For the second event we have 2/4 = 1/2, as we have two successful events (get a 4 or a 6) out of 4 possible events.
Then, we can calculate the probability as:
![P(7\&E)=P(7)\cdot P(E)=(1)/(4)\cdot(1)/(2)=(1)/(8)=0.125](https://img.qammunity.org/2023/formulas/mathematics/college/kw7dhrsruk5mx5t41ilebrdd7rbdleuiq2.png)
We can express this probability as a percentage by multiplying this result by 100%:
![P(7\&E)=0.125\cdot100\%=12.5\%](https://img.qammunity.org/2023/formulas/mathematics/college/relecqdkdhe11k2a5ag3mn86yq8o1ds6l8.png)
Answer: 12.5%