If 4-digits numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5 and 7, what is the probability of forming a number divisible by 5 when,
(i) the digits are repeated? (ii) the repetition of digits is not allowed?

Solution:

(i) When the digits are repeated:

Since 4-digit numbers greater than 5000 are randomly formed from the digits 0, 1, 3, 5 and 7, the leftmost digit is either 7 or 5. It means, the leftmost digit can be filled in 2 ways.

The remaining three places can be filled by any of the digits 0, 1, 3, 5 and 7, as repetition of digits is allowed. It means each of the remaining places can be filled in 5 ways.

Therefore,

• Total number of such 4-digit numbers greater than 5000 will be equal to
  2 × 5 × 5 × 5 - 1 = 250 - 1 = 249
  [Since 5000 cannot be counted, 1 is subtracted]
   
• A number is divisible by 5, if the digit at its units place is either 0 or 5. It means the rightmost digit can be filled in 2 ways.
  Total number of 4-digit numbers greater than 5000 that are divisible by 5 are
  2 × 5 × 5 × 2 - 1 = 100 - 1 = 99

Thus, the probability of forming a number divisible by 5 when the digits are repeated is 99/249 = 33/83

(ii) When repetition of digits is not allowed:

The thousands place can be filled with either of the two digits 5 or 7. i.e., it can be filled in 2 ways.

The remaining three places can be filled with any of the remaining four digits, but note that repetition is not allowed. So they can be filled in 4, 3, and 2 ways.

Hence, the total number of 4-digit numbers greater than 5000 will be

2 × 4 × 3 × 2 = 48

• When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining three digits.
  Therefore, number of 4-digit numbers starting with 5 and divisible by 5 are 3 × 2 = 6
   
• When the digit at the thousands place is 7, the units place can be filled in two ways, 0 or 5 and the tens and hundreds places can be filled with any two of the remaining three digits.
  Therefore, number of 4-digit numbers starting with 7 and divisible by 5 are 1 × 2 × 3 × 2 = 12

Hence,

Total number of 4-digit numbers greater than 50000 that are divisible by 5 are

6 + 12 = 18

Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is 18/43 = 3/8

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NCERT Solutions Class 11 Maths Chapter 16 Exercise ME Question 9

If 4-digits numbers greater than 5000 are randomly formed from the digits 0, 1, 3, 5 and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed?

Summary:

The probability of forming a number divisible by 5 when (i) the digits are repeated is 33/83 (ii) the repetition of digits is not allowed is 3/8