a veterinarian surveys her clients and finds that 32 percent of the households have dogs, 25 percent have…

a veterinarian surveys her clients and finds that 32 percent of the households have dogs, 25 percent have cats, and 11 percent have both dogs and cats. let event c be choosing a client who has cats and let event d be choosing a client who has dogs. which statements are true? check all that apply.\n$p(c|d)=0.78$\n$p(d|c)=0.44$\n$p(ccap d)=0.11$\n$p(ccap d)=p(dcap c)$\n$p(c|d)=p(d|c)$

a veterinarian surveys her clients and finds that 32 percent of the households have dogs, 25 percent have cats, and 11 percent have both dogs and cats. let event c be choosing a client who has cats and let event d be choosing a client who has dogs. which statements are true? check all that apply.\n$p(c|d)=0.78$\n$p(d|c)=0.44$\n$p(ccap d)=0.11$\n$p(ccap d)=p(dcap c)$\n$p(c|d)=p(d|c)$

Answer

Answer:

  • B. $P(D|C)=0.44$
  • C. $P(C\cap D)=0.11$
  • D. $P(C\cap D)=P(D\cap C)$

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Also, $P(A\cap B)$ represents the probability of both $A$ and $B$ occurring.

Step2: Calculate $P(D|C)$

We know that $P(C) = 0.25$, $P(D)=0.32$, and $P(C\cap D)=0.11$. Using the conditional - probability formula $P(D|C)=\frac{P(D\cap C)}{P(C)}$. Substituting the values, we get $P(D|C)=\frac{0.11}{0.25}=0.44$.

Step3: Identify $P(C\cap D)$

The problem states that 11 percent of households have both dogs and cats. So $P(C\cap D) = 0.11$.

Step4: Recall the property of intersection

The intersection of two events $A$ and $B$ is commutative, i.e., $P(A\cap B)=P(B\cap A)$. So $P(C\cap D)=P(D\cap C)$.

Step5: Check $P(C|D)$

$P(C|D)=\frac{P(C\cap D)}{P(D)}=\frac{0.11}{0.32}\approx0.34\neq0.78$. And since $P(C|D)=\frac{0.11}{0.32}\approx0.34$ and $P(D|C) = 0.44$, $P(C|D)\neq P(D|C)$.