Aptitude Discussion

Q. |
I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialling the first four, then what is the chance of dialling the correct number? |

✖ A. |
1/1001 |

✔ B. |
1/1000 |

✖ C. |
1/999 |

✖ D. |
1/990 |

**Solution:**

Option(**B**) is correct

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.

Hence required probability

\(=\left(\dfrac{1}{10}\right)^3\)

\(=\dfrac{1}{1000}\)

**RGSL**

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Even when we consider the repetition of numbers, each digit itself stands the chance to be any number i.e. 0 to 9

**Sherlock**

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there are 7 digits. Each digit can have a unique value ranging from 0-9 (i.e. total ten different possible values).

**Akshay**

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The questions suggests that the first four digits is dialed correctly, so neglect them. The last three digits is selected randomly from possible numbers 0 to 10. So, each position in the last three digits can take any number from 0 to 10 and also can be repetitive. That said the total number of possibilities for the last three positions is 10*10*10=1000.

So totally there are 1000 possible numbers and out of which only one number is correct. Therefore, P(correct number)=1/1000

Please correct in the previous post the possible numbers are from 0 to 9 not 10.

**Satyam**

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in question it is written only 7 digit no..how did u get 10???

Lets only talk about the last three digits which can be any number i.e. 0 to 9. Hence possible numbers are 10.

**Parul**

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from where did 10 came???

Why didn't you consider the the repetition of the numbers in the last three places lik 887, 999 etc