We have the cards numbered as 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.
Sample number is, n(S)=11.
We pick a card and want to get even number like 2, 4, 6, 8, 10, and 12.
Probability of picking an even number would be :-
P(even numbers) =
![(6)/(11)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/7wj1tgdjvap33ogn4smsaixko9mu8liosb.png)
Now we replace the card and want to get a 7 in second attempt.
Probability of picking a 7 would be :-
P(number 7) =
![(1)/(11)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ga4bowd2k08i0nubu8takxcjjncxr7u4jl.png)
Now Probability of this event = P(even numbers) × P(number 7)
Probability =
![((6)/(11) )((1)/(11) ) = (6)/(121)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/zuip7cxgifxln8ufgcybrzvpfezmjfcw0z.png)
So final answer is
.