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36% own a dog and .22 of those that own a dog also a cat. Furthermore .30 of people own a cat. What is the probability that a randomly selected family owns a dog and a cat?

We have two events and the intersection of those events. So we should look to use conditional probability between two events $E \& F$. Since we are asked to find the intersection we should use the formula $P(E \vert F) = \frac{P(EF)}{P(F)}$ and rearrange the equation to use the formula $P(EF)= P(E \vert F)P(F)$ Because of the way it is constructed and is worded as ".22 percent of those own that own a dog also own a cat." E should be the event that someone owns a dog and $P(E \vert F)=.22$ since 22% of those whom own a dog also own a cat.

We want the $P(EF)$ where

$E=$ the event someone owns a cat .30

$F=$ the event someone owns a dog .36

$E \vert F= .22$

$P(EF)= P(E \vert F)P(F)$

$P(E \vert F)=$ the event someone owns a cat given that they own a dog $= .22$ $P(F)=.36$

$P(EF) = P(E \vert F)P(F)$

$P(EF) = (.36)(.22)$

$P(EF) =.0792$ or $7.92$

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You have some faulty notation and contradictory numbers in your question, starting with "E|F = .22." But you are using the right fundamental equation.

I will use $C$ for 'owns cat', and $D$ for 'owns dog'. Then given $P(C) = 0.30,\,P(D) = 0.36,$ and $P(C|D) = 0.22,$ I will do a couple of the parts and hope that puts you on the right track for the rest.

  • $P(D \cap C) = P(D)P(C|D) = 0.36(0.22).$

  • $P(D|C) = P(D \cap C)/P(C) = P(D \cap C)/0.30,$ where the numerator is known from above.

Notes: (a) It is not correct that $P(C\cap D) = P(C)P(D)$ because events $C$ and $D$ are not independent; do you know how to show that? (b) This not a good problem for Venn Diagrams, because conditional probabilities are not easily represented by them.