Question

--A software company uniformly randomly generates 14-character encryption keys from 26 letters and 10 digits. Repeated characters are allowed, and the order of the characters matters.

(a) How many different keys can be generated, if there are no restrictions at all on the keys?

(b) What is the probability that a key consists of only letters?

(c) What is the probability that a key starts and ends with a digit?

(d) If the digits are removed from the 14-character key and all spaces are removed, what is the probability that the resulting string of letters is AQRRFHMK?

(e) What is the probability that exactly 4 of the 14 characters are letters?

(f) What is the probability that all of the characters are different?

(g) What is the probability that there are at least two F’s?

--the Binomial Theorem

(a) What is the coefficient of x5y9 in (x + 2y)14?

(b) What is the coefficient of x8 in (3x ? 4)17?

-- For a group of k people, let Ak be the event “no two people have the same birthday norbirthdays one day apart”. As usual, ignore leap years, and assume all birthdays are equally likely. Also note that Dec 31 and Jan 1 are considered one day apart.

For each k, Find a lower bound on P(Ak); that is, find a number ak ? (0,1) such that P(Ak) ? ak, and explain your reasoning.

-- For a poker hand with k cards, a flush means all cards have the same suit, and a straightmeans all cards are in sequence according to rank, and a straight flush means both at the same time. A straight may start with an ace, A-2-3-4(-5), or may end with an ace, (10)-J-Q-K-A, but you can’t have a “wraparound” straight, e.g. Q-K-A-2-3 is not considered a straight.

For the two cases k = 4 and k = 5, for a hand with k cards, find the probabilities of a hand dealt uniformly at random having (a) a straight flush, (b) a (non-straight) flush, and (c) a (non-flush) straight, and for each value of k, rank these three hands in order of probability. For four-card poker (the case k = 4), which hand should be considered “stronger” – a flush or a straight?

Answer 1

(a)

Total number of characters: 26+10= 36

Since there is no restrictions and characters can be repeated so possible number of different keys is

(b)

Now we need to find the number of keys with only letters. Since there are 26 letters so possible number of keys having only letters is

The probability that a key consists of only letters is

(c)

Now we need to find the number of keys with start and end with digits. First space of the key can be filled by 10 digits and last space can also be filled by 10 digits. Rest 12 space can be filled by 36 charatcters each so possible number of keys start and end with digits is

So the probability that a key starts and ends with a digit is

(d)

We need to fill 8 spaces out of 14 spaces by AQRRFHMK in the same order. Number of ways of choosing rest 6 positions out of 14 is C(14,6) = 3003

These 6 spaces can be filled in 10 ways each so possible number of keys having AQRRFHMK is

The required probability is