In a relay race there are five teams A, B, C, D and E. (a) What is the probability that A, B and C finish first, second and third, respectively. (b) What is the probability that A, B and C are first three to finish (in any order) (Assume that all finishing orders are equally likely)

Step 1: Calculate total possible finishing orders

We have 5 teams (A, B, C, D, E) participating in a relay race. The total number of ways these 5 teams can finish the race is the number of permutations of 5 distinct items, which is calculated as $5!$ (5 factorial).

Step 2: Calculate probability for part (a)

For part (a), we want the probability that A finishes first, B second, and C third. There is only 1 way for A to be first, 1 way for B to be second, and 1 way for C to be third. The remaining 2 teams (D and E) can finish in $2!$ ways in the 4th and 5th positions. So, the number of favorable outcomes is $1 \times 1 \times 1 \times 2! = 2$.

Step 3: Calculate probability for part (b)

For part (b), we want the probability that A, B, and C are the first three to finish, in any order. The number of ways A, B, and C can occupy the first three positions is $3!$ (permutations of A, B, C). The remaining 2 teams (D and E) can finish in $2!$ ways in the 4th and 5th positions. So, the number of favorable outcomes is $3! \times 2! = 6 \times 2 = 12$.

Step 4: Simplify probabilities

Finally, we simplify the probabilities calculated in the previous steps to their simplest fractional forms.

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