Question

a code is 4 random letters and 3 random digits. what is the probability of the 4 letters all being different while the 3. numbers at the end are in ascending order?

b) in a group of 10 codes.... what is the probability that at least two have all different letters and numbers in ascending order

c) in 1000 codes what is the wxpected value and standard deviation for the codes with all different letters and ascending digits

Answer 1

The code is formed using 4 random letters and 3 random digits.

We know,

The probability of an event = (the number of favorable cases to that event)/(Total number of outcomes)

We have to find the probability of the 4 letters all being different while the 3 numbers at the end are in ascending order.

Here, total number of outcomes = (number of arrangements of 4 letters from 26 alphabets where each alphabet can repeat upto 4 times) x (number of arrengements of 3 digits from 10 digits where each digit can repeat upto 3 times)

= 264 x 103

And the number of favorable cases to the event = (number of arrangements of 4 different letters from 26 alphabets where repetition is not allowed) x (number of arrengements of 3 digits in ascending order out of 10 digits)

Now, for the number of arrengements of 3 digits in ascending order out of 10 digits we consider the following cases:

1st digit 2nd digit number of cases

0 1 ... = 8

0 2 ... = 7

...

0 8 ... = 1

1 2 ... = 7

1 3 ... = 6

...

1 8 ... = 1

....

....

7 8 ... = 1

Therefore, the number of numbers that start with 0 = (8 + 7 + ... + 1) = 36

the number of numbers that start with 1 = (7 + 6 + ... + 1) = 28

the number of numbers that start with 2 = (6 + 5 + ... + 1) = 21

and so on.

Hence, the number of arrengements of 3 digits in ascending order out of 10 digits = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

So, the number of favorable cases to the event = 26P4 x 120

Therefore, the required probability = (26P4 x 120)/264 x 103 = 0.0942

Answer: The probability of the 4 letters all being different while the 3 numbers at the end are in ascending order is 0.0942.

b) Let X = random variable denoting the number codes out of 10 which have all different letters and numbers in ascending order

Since each group is independent of the other and the probability of having all different letters and numbers in ascending order 0.0942 remains same for all groups

X ~ bin(n = 10, p = 0.0942)

The p.m.f. of X is given by,

Then, the probability that at least two have all different letters and numbers in ascending order is

= 1 - (0.3718 + 0.3866)

= 0.2416

Answer: in a group of 10 codes, the probability that at least two have all different letters and numbers in ascending order is 0.2416.

c) For 1000 codes, the Poisson approximation to binomial will be applicable due to the following conditions:

i) Here n = 1000 is very large i.e.

ii) p = 0.0942 is very small i.e.

iii) np = 94.2 is finite

i.e. X ~ Poisson ( = np = 94.2)

For Poisson distribution E(X) = Var (X) = = 94.2

Therefore, s.d.(X) = = 9.71 (rounded to 2 decimal places)

Answer: In 1000 codes, the expected value and standard deviation for the codes with all different letters and ascending digits are 94.2 and 9.71 respectively.