Question

Question 10 The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., 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?

Class X1 - Maths -Probability Page 409

Answer 1

There are 10 digits out of which we have to choose only 4 and no digit is repeated.
Thus we will apply :-
$p \binom{10}{4}$
=> 10!/(10-4)!
=> 10!/6! = 10×9×8×7×6!/6!
=> 5040

Hence total outcomes = 5040

Now lock will by one correct sequence hence
probability = 1/5040

Answer 2

Here ,repetition is not allowed .
so, first place can be filled in 10 ways
2nd place can be filled in (10 - 1) = 9 ways
3rd place can be filled in ( 10 - 2) = 8 ways
and 4th place can be filled in ( 10 - 3) = 7 ways

Hence, by fundamental principal of counting ,
total number of ways = 10 × 9 × 8 × 7
= 5040 ways .

we know, in which only 1 sequence can open the lock.
so, number of favourable cases n( E ) = 1
total number of possible cases n( S ) = 5040

so, P ( E ) = n( E )/n( S ) = 1/5040