Question

question 1 (1 point)
four puppies are selected at random for a commercial from a dog breeder that has 26 puppies all of different weights in grams. what is the probability that they will be in order of weight?
(a) \frac{4!}{_{26}p_{4}}
(b) \frac{1}{26^{4}}
(c) \frac{4}{26!}
(d) \frac{1}{_{26}p_{4}})

Explanation:

Step1: Calculate number of permutations of 4 puppies

The number of ways to select and order 4 puppies out of 26 is given by the permutation formula $_{n}P_{r}=\frac{n!}{(n - r)!}$, where $n = 26$ and $r=4$, so $_{26}P_{4}=\frac{26!}{(26 - 4)!}=\frac{26!}{22!}=26\times25\times24\times23$.

Step2: Determine favorable arrangements

There are only 2 possible ordered arrangements (ascending or descending) out of all the possible arrangements of the 4 - selected puppies. The total number of arrangements of the 4 selected puppies is $4!$. But we are interested in only 2 (either increasing or decreasing weight order) out of these $4!$ arrangements of the 4 puppies. The number of favorable arrangements of the 4 - selected puppies in order of weight is 2. The probability $P$ of an event is the number of favorable outcomes divided by the number of total outcomes. The total number of ways to pick and order 4 puppies out of 26 is $_{26}P_{4}$, and the number of favorable ways (in order of weight) is 2. Since we are only considering one - type of order (either increasing or decreasing), the number of favorable arrangements is 1. So the probability $P=\frac{1}{_{26}P_{4}}$.

Answer:

D. $\frac{1}{_{26}P_{4}}$