If you take two balls at a time, there are always * * * possibilities C (2,6) =15. Without 5 and 6, there is a possibility C (2,4) = 6, so the probability of 5 or 6 is 1- 6/ 15= 3/5.
There may be five kinds of 5, namely 1 and 5, 2 and 5, 3 and 5, 4 and 5, 6 and 5, so the probability of the number 5 is 5/ 15= 1/3.
2. Touch the ball back twice, one ball at a time. The probability that the number 5 or 6 appears is 5/9, the probability that the number 5 appears is 1 1/36, and the probability that the sum of the numbers is 5 or 6 is 1/4.
Touch the ball back twice, and there are always 6×6= 36 possibilities.
The probability that 5 and 6 do not appear is 4×4= 16, so the probability that 5 or 6 appears is 1- 16/36= 5/9.
There are 5×5= 25 possibilities for not appearing 5, so the probability of appearing 5 is1-25/36 =11/36.
There are four possibilities for digital sum to be 5: 1 and 4, 2 and 3, 3 and 2, 4 and 1 * *, and five possibilities for digital sum to be 6: 1 and 5, 2 and 4, 3 and 3, 4 and 2, 5 and 1 * *. So the sum of the numbers is 5 or 6***, and there are nine possibilities.
Touch the ball twice, don't put it back, one ball at a time. The probability that the number 5 or 6 appears is 3/5, the probability that the number 5 appears is 1/3, and the probability that the sum of the numbers is 5 or 6 is 4/ 15.
This problem is the same as the variant 1.
By the way, the multiple-choice question above is C.
I hope you can adopt it, but you can ask if you don't understand it. Thank you.