A bag contains four tickets numbered 00,01,10,11. Four tickets are chosen at random with replacement, the probability that sum of the numbers on the tickets is 23 is:

3/32

Sum of numbers = 23 → 3’s in ones place
                                    2’s in tens place

Possibilities

11 + 11 + 01 + 00

11 + 10 + 01 + 01

Total no. of ways to get two 11, one 01, one 00 → no. of arrangements = $\frac{4!}{2!}$

One 11, one 10, two 01 → $\frac{4!}{2!}$

Total no. of ways = 4⁴

Probability = $\frac{\frac{4!}{2!}+\frac{4!}{2!}}{4^4}=\frac{3}{32}$