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The number lock of a suitcase has four wheels,each labelled with 10 digits (ie) from 0 to 9.The lock opens with a sequence of four digits with no repeats.What is the probability of a person getting the right sequence to open the suitcase?

$\begin{array}{1 1}(A)\;\large\frac{1}{5040}\\(B)\;\large\frac{2}{4035}\\(C)\;\large\frac{3}{5040}\\(D)\;\text{None of these}\end{array} $

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1 Answer

  • Required probability =$\large\frac{n(E)}{n(S)}$
Step 1:
Given repetition is not allowed.
$\therefore$ First place can be filled in 10 ways.
Second place can be filled in 9 ways.
Third place can be filled in 8 ways
Fourth place can be filled in 7 ways
$\therefore$ Total number of ways =$10\times 9\times 8\times 7$
$\Rightarrow 5040$ ways
Step 2:
Total number of outcomes n(S)=5040
Only one sequence will be right which will open the lock
No of favorable cases n(E)=1
Required probability=$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{1}{5040}$
Hence (A) is the correct answer.
answered Jul 7, 2014 by sreemathi.v

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