(b) How many ways can you arrange the letters in the word STATISTICS? If all ar- rangements are equally likely, what is the probabilitiy the two 'i's are next to each other. (c) There are 3 arrangements of the word DAD, namely DAD, ADD, and DDA. How many arrangements are there of the word PROBABILITY? (d) In a ballroom dancing class the students are divided into group \( A \) and group \( B \). There are six people in group \( A \) and seven in group \( B \). If four \( A \) s and four \( B \mathrm{~s} \) are chosen and paired off, how many pairings are possible? (e) Let \( A \) and \( B \) be two events. Suppose the probability that neither \( A \) or \( B \) occurs is \( 2 / 3 \). What is the probability that one or both occur? (f) Let \( C \) and \( D \) be two events with \( P(C)=0.25, P(D)=0.45 \), and \( P(C \cap D)=0.1 \). What is \( P\left(C^{c} \cap D\right) \) ?

The arrangements of the letters in the word "STATISTICS" can be computed using the factorial formula, considering the repeated letters. The word "STATISTICS" consists of 10 letters with specific frequencies: S (3), T (3), A (1), I (2), C (1). The total number of distinct arrangements is calculated as \( \frac{10!}{3! \cdot 3! \cdot 2! \cdot 1! \cdot 1!} = 16800 \). To find the probability that the two 'i's are next to each other, we can treat the two 'i's as a single unit, giving us the letters: "SSTTACCI" which includes 9 units. The arrangements become \( \frac{9!}{3! \cdot 3! \cdot 1! \cdot 1!} = 8400 \). Thus, the probability is \( \frac{8400}{16800} = \frac{1}{2} \). When arranging the letters in the word "PROBABILITY", we take into account the counts of each letter: P (1), R (1), O (1), B (2), A (1), I (1), L (1), T (1), Y (1). This leads to a total arrangement count of \( \frac{11!}{2!} = 19958400 \) due to the two 'B's. To calculate the number of pairings between groups \( A \) and \( B \), first choose 4 people from each group. The combinations for choosing 4 from 6 in group \( A \) is \( \binom{6}{4} = 15 \) and from group \( B \) is \( \binom{7}{4} = 35 \). After selection, the number of ways to pair is \( 4! = 24 \). Therefore, the total number of pairings is \( 15 \times 35 \times 24 = 12600 \).