Question

1. game show the people on the list at the right will be considered to participate in a game show. what is the probability that wyatt, gabe, and isaac will be chosen as the first three contestants?

day 1 standings
mcafee, david
ford, gabe
standish, tristan
nochols, wyatt
purcell, jack
anderson, bill
wright, isaac
filbert, mitch

Explanation:

Step1: Count total number of people

From the list, we can see there are 8 people (David, Gabe, Tristan, Wyatt, Jack, Bill, Isaac, Mitch). So total number of people \( n = 8 \).

Step2: Calculate number of permutations of 3 people

The number of permutations of choosing 3 people out of 8 is given by the permutation formula \( P(n,r)=\frac{n!}{(n - r)!} \), where \( n = 8 \) and \( r=3 \). But we can also think of it as: the number of ways to choose the first contestant is 8, the second is 7 (since one is already chosen), and the third is 6. So total number of ways to choose first 3 contestants is \( 8\times7\times6=336 \).

Step3: Calculate number of favorable permutations

We want Wyatt, Gabe, and Isaac to be chosen as the first three. The number of permutations of these three people among themselves is \( 3! = 3\times2\times1=6 \) (since Wyatt, Gabe, Isaac can be arranged in 3! ways as first, second, third).

Step4: Calculate probability

Probability \( P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{3!}{8\times7\times6}=\frac{6}{336}=\frac{1}{56} \).

Answer:

\(\frac{1}{56}\)