Question

Dialing a telephone number, an old person forgets last three digits. Remembering only that these digits are different, he dialed at random. What is the chance that the number dialed is   CORRECT?

1. \(\frac{1}{1000}\)
2. \(\frac{1}{720}\)
3. \(\frac{9}{70}\)
4. \(\frac{1}{7840}\)

Answer 1

Correct Answer - Option 2 : \(\frac{1}{720}\)

Given:

The person forgot last three digits

Concept Used:

Probability = Number of desire event / Number of total event

Calculation:

Total number of digits = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} = 10 

We have to fill 3 places

⇒ Total possible cases = ¹⁰C₃ × 3!

⇒ \({10! \over {3! \times (10 - 3)!}} \times 3!\)

⇒ \({{10 \times 9 \times 8} \over {3 \times 2 \times 1}} \times 3 \times 2 \times 1\)

⇒ 720

Possibility of correct the number is 1/720