Question

A five-digit number is written down at random. The probability that the number is divisible by $5$ and no two consecutive digits are identical, is:

• A. $\frac{1}{5}$
• B. $\frac{34}{90}.{\left(\frac{9}{10}\right)}^3$
• C. ${\left(\frac{3}{5}\right)}^4$
• D. ${\left(\frac{2}{5}\right)}^4$

Answer 1

The correct option is B $\frac{34}{90}.{\left(\frac{9}{10}\right)}^3$

The possible option for the last digits are $0,5$ only. ($2$ out of $10$)
Therefore number of possible options for second last digit are $9$ out of $10($can't be equal to last digit so $10-1)$
Similarly, number of possible options for third last digit are $9$ out of $10;$
In Fourth last digit
$\left(i\right)$ If $0$ comes at fourth last digit, then number of possible options for first digit are $9$ and out of them any $9$ options can come in first digit.
$\left(ii\right)$ If $0$ does not come at fourth last digit, then number of possible options are $9$ and only $8$ options can come at first digit.
$\therefore$ Required probability =$(\frac{9}{9}\times \frac{9}{10}\times \frac{9}{10}\times \frac{9}{10}\times \frac{2}{10})+(\frac{8}{9}\times \frac{9}{10}\times \frac{9}{10}\times \frac{9}{10}\times \frac{2}{10})$
$=\frac{17}{9}\times {\left(\frac{9}{10}\right)}^3\times \frac{2}{10}=\frac{34}{90}\times {\left(\frac{9}{10}\right)}^3$