Question

Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is

Options

•  \[\frac{1}{5}\]

   

•  \[\frac{4}{5}\]

   

• \[\frac{1}{30}\]

   

•  \[\frac{5}{9}\]

   

Answer 1

The given digits are 0, 2, 3 and 5.

_____	_____	_____	_____
Thousands	Hundreds	Tens	Ones

Now, there are 3 ways to fill the thousands place (0 cannot occupy the thousands place), 3 ways to fill the hundreds place, 2 ways to fill the tens place and 1 way to fill the ones place.
Total number of four digit numbers formed = 3 × 3 × 2 × 1 = 18
We know that a number is divisible by 5 if it ends in 0 or 5.
When 0 is at the ones place,
Number of four digits numbers divisible by 5 formed = 3 × 2 × 1 = 6
When 5 is at the ones place,
Number of four digits numbers divisible by 5 formed = 2 × 2 × 1 = 4                    

(0 cannot occupy the thousands place)        
Total number of four digit numbers divisible by 5 = 6 + 4 = 10
∴ P(four digit number formed is divisible by 5) = \[\frac{\text{ Total number of four digit numbers divisible by 5 } }{\text{ Total number of four digit numbers formed} } = \frac{10}{18} = \frac{5}{9}\]