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:
Answer & explanation
Correct answer: option 1
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 = $\frac{\frac{4!}{2!}+\frac{4!}{2!}}{4^4}=\frac{3}{32}$