Take a chance with the padlock

1 thing I would suggest to mention that a unit can mean that 0 & 9 could be together..I didnt take that combination in first attempt....

Now, digits that can be together are 0 & 1, 1 & 2, 2 & 3,......8 & 9, 9 & 0

Say we can have 0s & 1s, we can have 2^4 =16 combinations.

For 1s & 2s we can have 16-1=15 combinations (as 1111 is already counted above).

Similarly till 0s & 9s where we will count 14 cases only(as both 0000 & 9999 have been already counted in above cases).

So total 150 cases possible.

P(1st attempt)=150/10000

P(2nd attempt) =9850/10000 x 150/9999

P(3rd attempt) =9850/10000 x 9849/9999 x 150/9998

Total Probability of success=0.044

Firstly, let's consider how many possibilities there are:

Let the first digit be 0. Then there are 2 cases:

1. The remaining three digits are either 1 or 0. In which case, we have $2^3=8$ possibilities.

2. The remaining digits are either 9 or 0. In which case we also have $2^3=8$ possibilities.

But here, we counted 0000 twice. So when the first digit is 0, we have $2 \times 8 -1=15$ possibilities.

Now, we get identical cases for when the first digit varies. There are 10 possibilities for the first digit, hence, in total, we have $10 \times 15=150$ possible padlock configurations which satisfy the above conditions.

Now there are $10^4=10000$ different combinations in total. Here it is easier calculating the probability that all 3 attempts fail:

1st attempt: we have $P_1$ (fail) $=\frac{10000-150}{10000}=\frac{9850}{10000}$

2nd attempt: Since the first attempt failed, we have 1 less possibility. Then $P_2$ (fail) $=\frac{9850-1}{10000-1}=\frac{9849}{9999}$

3rd attempt: Since the first and second attempt failed, we have 2 less possibilities. Then $P_3$ (fail) $=\frac{9850-2}{10000-2}=\frac{9848}{9998}$

So the probability of failing all is

$\frac{9850}{10000} \times \frac{9849}{9999} \times \frac{9848}{9998}=0.9556672...$

So the probability of succeeding is $1-0.9556672...=0.044332741...=0.0443(3sf)$