Two letters are chosen at random *with replacement *from the word PROBABILITY. Note there are 11 letters in total in PROBABILITY.

a) P and T each only occur once in the word. The probability of choosing P and T in a particular order (eg P then T) is

Pr(PT) = Pr(P) x Pr(T) (as they are *independent *events)

= (1/11) x (1/11) = 1/121

There are only two orders the letters can be in: PT or TP, so the probability of P and T in any order is

Pr(PT) + Pr(TP) = 1/121 + 1/121 = 2/121

b) The probability of choosing 2 B's is

(2/11) x (2/11) = 4/121

There is only one order for choosing these.

The probability of choosing two different letters is 1 - the probability of choosing two letters that are the same. For all of the letters (except the B's and I's, which are repeated twice), the probability of choosing the same one twice is

(1/11) x (1/11) = 1/121

Therefore the probability of choosing two letters that are the same is

Pr(two of any letter except B or I) + Pr(two B's or two I's) = 7 x (1/121) + 2 x (4/121) = 17/121

The probability of choosing two different letters is then 1 - 15/121 = 106/121.

The probability, then, that the letters are both B's *or *they are different is given by

Pr(two B's or two different letters) = 4/121 + 106/121 = 110/121

c) The probability that a B or an I is chosen is

Pr(B or I) = 2/11 + 2/11 = 4/11

If another is letter is chosen, the probability that that is neither a B nor an I is

Pr(not B and not I) = 7/11

Combining these two *independent* events, the probability that one of the letters chosen is a B or an I and the other letter is neither a B nor a I is given by

Pr(B or I *and* not B and not I) = Pr(B or I) x Pr(not B and not I) = 4/11 x 7/11 = 28/121

**Answer**

**a) i) 1/121 ii) 2/121 (now correct...)**

**b) 110/121**

**c) 28/121**

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