Question

A typical PIN (personal identification number) is a sequence of any four symbols chosen from the 26 letters the alphabet and the ten digits. If all PINs are equally likely, what is the probability that a randomly chosen PIN contains repeated symbols?

Answer 1

We can find the total number of possible PINs by multiplying the total number of symbols 4 times. Then we can find the number of PINs without repeated symbols using permutations ${}^{36}{P_4}$. Then we can find the number of PINs that contain repeated symbols by subtracting the number of PINs without repeated symbols from the total number of PINs. Now we can find the required probability by dividing the number of PINs with symbols repeating with the total number of PINs.

Complete step by step answer:

We need to make a PIN with 4 symbols.
We have to choose the symbols from 26 alphabets and 10 digits. So total number of symbols is given by, $26 + 10 = 36$ .
Now we can find the total number of possible PINs. We need to take 4 symbols and each symbol can be any of the 36 symbols.
Therefore, the number of possible PIN is $36 \times 36 \times 36 \times 36 = 1,679,616$ .
Now we can find the number of PINs without symbols repeating. We need to select 4 symbols from 36 symbols. It is given by the permutation ${}^{36}{P_4}$.
We know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ .
$\Rightarrow{}^{36}{P_4} = \dfrac{{36!}}{{\left( {36 - 4} \right)!}}$
On simplifying the factorial, we get,
$\Rightarrow{}^{36}{P_4} = 36 \times 35 \times 34 \times 33$
After doing the multiplication, we get,
$\Rightarrow{}^{36}{P_4} = 1,413,720$
Therefore, the number of PIN without repeating symbols is 1413720.
Now we can find the number of PINs that contain repeated symbols. It is given by,
$$Number{\text{ }}of{\text{ }}PIN{\text{ }}containing{\text{ }}repeated{\text{ }}symbol{\text{ }} = {\text{ }}total{\text{ }}number{\text{ }}of{\text{ }}PINs{\text{ }} - {\text{ }}number{\text{ }}of{\text{ }}PINs{\text{ }}without{\text{ }}repeating{\text{ }}symbols.$$$ = 1679616 - 1413720$$ = 265896$. We know that the probability of an event is given by the number of favourable outcomes divided by the total number of outcomes.$P\left( {random{\text{ }}PIN{\text{ }}has{\text{ }}repeated{\text{ }}symbol} \right) = \dfrac{{no.{\text{ }}of{\text{ }}PINs{\text{ }}having{\text{ }}repeated{\text{ }}symbol}}{{total{\text{ }}number{\text{ }}of{\text{ }}PINs}}$On substituting the values, we get,$P = \dfrac{{265896}}{{1679616}}$After the division we get,$P = 0.158$
Therefore, the required probability is 0.158.

Note: An alternate solution to this problem is,
We have the number of possible PIN as $36 \times 36 \times 36 \times 36 = 1,679,616$ .
We also have the number of PIN without repeating symbols as ${}^{36}{P_4} = 36 \times 35 \times 34 \times 33 = 1,413,720$
Then we can find the probability that PIN selected has no repeating symbol
$P\left( {random{\text{ }}PIN{\text{ }}has{\text{ }}no{\text{ }}repeated{\text{ }}symbol} \right) = \dfrac{{no.{\text{ }}of{\text{ }}PINs{\text{ }}having{\text{ }}no{\text{ }}repeated{\text{ }}symbol}}{{total{\text{ }}number{\text{ }}of{\text{ }}PINs}}$
On substituting the values, we get,
$P = \dfrac{{1413720}}{{1679616}}$
After the division we get,
$P = 0.842$
Then the probability that PIN selected has repeating symbol is given by,
$P\left( {random{\text{ }}PIN{\text{ }}has{\text{ }}repeated{\text{ }}symbol} \right) = 1 - P\left( {random{\text{ }}PIN{\text{ }}has{\text{ }}no{\text{ }}repeated{\text{ }}symbol} \right)$
$= 1 - 0.842$
$= 0.158$
Therefore, the required probability is 0.158