Question

the eau claire humane society currently has 18 dogs available to adopt. 10 are mutts and the other 8 are purebreeds. if a volunteer randomly selects 2 different dogs to walk... a. what is the probability that they are both purebreeds? b. what is the probability that one is a purebreed and the other is a mutt? note: you can earn partial credit on this problem. preview my answers submit answers your score was recorded. your score was successfully sent to canvas. you have attempted this problem 2 times. you received a score of 0% for this attempt. your overall recorded score is 0%. you have unlimited attempts remaining.

Explanation:

Step1: Calculate total number of ways to select 2 dogs

The number of ways to choose 2 dogs out of 18 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 18$ and $r=2$. So $C(18,2)=\frac{18!}{2!(18 - 2)!}=\frac{18\times17}{2\times1}= 153$.

Step2: Calculate number of ways to select 2 pure - breeds

The number of ways to choose 2 pure - breeds out of 8 is $C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7}{2\times1}=28$.

Step3: Calculate probability of selecting 2 pure - breeds

The probability $P(\text{both pure - breeds})=\frac{C(8,2)}{C(18,2)}=\frac{28}{153}\approx0.183$.

Step4: Calculate number of ways to select 1 pure - breed and 1 mutt

The number of ways to choose 1 pure - breed out of 8 and 1 mutt out of 10 is $C(8,1)\times C(10,1)$. Using the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, $C(8,1)=\frac{8!}{1!(8 - 1)!}=8$ and $C(10,1)=\frac{10!}{1!(10 - 1)!}=10$. So $C(8,1)\times C(10,1)=8\times10 = 80$.

Step5: Calculate probability of selecting 1 pure - breed and 1 mutt

The probability $P(\text{1 pure - breed and 1 mutt})=\frac{C(8,1)\times C(10,1)}{C(18,2)}=\frac{80}{153}\approx0.523$.

Answer:

a. $\frac{28}{153}$
b. $\frac{80}{153}$