Question

Question :

Out of 6 digit numbers formed by using digits 0, 1, 2, 3, 4, 5 without repetition , one is selected at random then the $\bf{Probability}$ that the selected number is divisible by 25 is ?​

Answer 1

Given:

6 digit numbers are formed using the digits 0, 1, 2, 3, 4 and 5 without repetition.

To find:

The probability that the selected number is divisible by 25.

Solution:

Total 6 digit numbers that can be formed are: 5*5*4*3*2*1 = 5*5! = 600

(The number should not start with 0)

To be divisible by 25 the number must end with 00 or 25 or 50 or 75,

In this case the number can only end with 50.

So fixing the last two digits as 50, we have 4 places left to be filled with 4 digits, hence the number of ways are 4*3*2*1 = 4!

The required probability will be 24/ 600 = 0.04

Therefore the required probability is 0.04.

Answer 2

Given : Out of 6 digit numbers formed by using digits 0, 1, 2, 3, 4, 5 without repetition , one is selected at random

To Find : Probability that the selected number is divisible by 25

Solution:

6 digit numbers formed by using digits 0, 1, 2, 3, 4, 5 without repetition

1st digit can not be 0

Hence can be selected in 5 ways

rest 5 can be arranged in 5! ways

= 5 * 5!  =  600   numbers are possible

selected number is divisible by 25   if it end with  00 , 25 , 50 or 75

00 and 75  is not possible  as only one zero is there and there is no 7

Case 1:  End  with    25

remaining digits  0 , 1 , 3 , 4

possible  = 3*3!   = 18

Case 2:  End  with    50

remaining digits  1 , 2 , 3 , 4

possible  =4!   = 24

Hence 18 + 24 = 42  numbers are divisible by 25

Probability that the selected number is divisible by 25  = 42/600

= 7/100

= 0.07

0.07 is the Probability that the selected number is divisible by 25

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