Answer :
Answer:
The probability that the pet seen was sick is 53.57%.
Step-by-step explanation:
The probability of an event E is the ration of the number of favorable outcomes to the total number of outcomes.
[tex]P(E)=\frac{n(E)}{N}[/tex]
Here,
n (E) = number of favorable outcomes
N = total number of outcomes
Denote the events as follows:
C = the pet is a cat
D = the pet is a dog
O = the pet is some other animal
S = the pet is sick.
The data provided is summarized as follows:
n (C) = 28
n (C ∩ S) = 3
n (D) = 42
n (B ∩ S) = 4
n (O) = 24
n (O ∩ S) = 8
Compute the probability that a cat was sick as follows:
[tex]P(C\cap S)=\frac{n(C\cap S)}{n(C)}=\frac{3}{28}[/tex]
Compute the probability that a dog was sick as follows:
[tex]P(D\cap S)=\frac{n(D\cap S)}{n(D)}=\frac{4}{42}=\frac{2}{21}[/tex]
Compute the probability that another animal was sick as follows:
[tex]P(O\cap S)=\frac{n(O\cap S)}{n(O)}=\frac{8}{24}=\frac{1}{3}[/tex]
Compute the probability that the pet seen was sick as follows:
P (S) = P (C ∩ S) + P (D ∩ S) + P (O ∩ S)
[tex]=\frac{3}{28}+\frac{2}{21}+\frac{1}{3}\\=\frac{9+8+28}{84}\\=\frac{45}{84}\\=0.5357[/tex]
Thus, the probability that the pet seen was sick is 53.57%.