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)
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Step-by-Step Solution
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 1×1×1×2!=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!×2!=6×2=12.
Step 4: Simplify probabilities
Finally, we simplify the probabilities calculated in the previous steps to their simplest fractional forms.