Question

The number lock of a suitcase has $4$ wheels each labelled with ten digits i.e. from $0$ to $9.$ What is the probability of a person getting the right sequence to open the suitcase, if
$A:$ Repetition is not allowed.
$B:$ Repetition is allowed.

• A. $P\left(A\right)=\frac{1}{5040}$
• B. $P\left(A\right)=\frac{1}{1080}$
• C. $P\left(B\right)=\frac{1}{{10}^5}$
• D. $P\left(B\right)=\frac{1}{{10}^4}$

Answer 1

Answer: (A), (D)

$P\left(B\right)=\frac{1}{{10}^4}$

Number of wheels in lock of suitcase $=4$

$A:$ Since repetition is not allowed.
Now first wheel can take number from $0$ to $9.$
Second wheel can have any of the remaining $9$ digits.
Similarly, Third wheel can have any of the remaining $8$ digits.
Similarly, Fourth wheel can have any of the remaining $7$ digits.
So the number of four digit lock that can be formed without repititon $=10\times 9\times 8\times 7=5040$
But, lock can be opened with only one of all these numbers.
$\therefore P\left(A\right)=\frac{1}{5040}$

$B:$ Since repetition is allowed.
All fours wheels can take all $10$ digits.
So the number of four digit lock that can be formed with repetition $={10}^4$
But, lock can be opened with only one of all these numbers.
$\therefore P\left(B\right)=\frac{1}{{10}^4}$