Question
A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:
  1. $\frac{1}{5}$
  2. $\frac{1}{5}\big(\frac{9}{10}\big)^3$
  3. $\big(\frac{3}{5}\big)^4$
  4. $\text{None of these}$

Answer

  1. $\text{None of these}$
Solution:
If last digit is either O or 5 then the number is divisible by 5.
Case : 1
Last digit is 0.
First three places can be selected by 9 × 9 × 9 = 729 ways.
If c = 0 then three places can be selected by 9 × 8 × 1 = 72
If C ≠ 0 then 729 - 72 = 657
Fourth place has 8 choices = 657 × 8 = 5256
Total = 72 + 5256 = 5904
Case : 2
If C = 5
First place other than 5
then first three places can be filled in 8 × 8 × 1 = 64
If first place is 5 then first three places can be filled in 1× 9 × 1 = 9 ways.
If third place is other than 5 then 729 - 64 - 9 = 656 ways.
For fourth place has 8 choices.
As per required condition = (64 + 9) × 9 + 656 × 8 = 5905
required probability $=\frac{5904+5905}{9\times10\times10\times10\times10}=\frac{11809}{90000}$
NOTE: Answer not matching with back answer.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free