finding conditional probabilities using a venn diagram\nuse the venn diagram to calculate conditional…

finding conditional probabilities using a venn diagram\nuse the venn diagram to calculate conditional probabilities.\nwhich conditional probabilities are correct? check all that apply.\n$p(d | f)=\frac{6}{34}$\n$p(e | d)=\frac{7}{25}$\n$p(d | e)=\frac{7}{25}$\n$p(f | e)=\frac{8}{18}$\n$p(e | f)=\frac{13}{21}$

finding conditional probabilities using a venn diagram\nuse the venn diagram to calculate conditional probabilities.\nwhich conditional probabilities are correct? check all that apply.\n$p(d | f)=\frac{6}{34}$\n$p(e | d)=\frac{7}{25}$\n$p(d | e)=\frac{7}{25}$\n$p(f | e)=\frac{8}{18}$\n$p(e | f)=\frac{13}{21}$

Answer

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In terms of the Venn - diagram, $P(A|B)=\frac{n(A\cap B)}{n(B)}$, where $n(A\cap B)$ is the number of elements in the intersection of $A$ and $B$, and $n(B)$ is the number of elements in $B$.

Step2: Calculate $P(D|F)$

$n(D\cap F)=7 + 1=8$, $n(F)=7 + 1+21 + 5=34$. So $P(D|F)=\frac{n(D\cap F)}{n(F)}=\frac{8}{34}=\frac{4}{17}\neq\frac{6}{34}$.

Step3: Calculate $P(E|D)$

$n(E\cap D)=6 + 1=7$, $n(D)=13 + 6+1 + 5=25$. So $P(E|D)=\frac{n(E\cap D)}{n(D)}=\frac{7}{25}$.

Step4: Calculate $P(D|E)$

$n(D\cap E)=6 + 1=7$, $n(E)=4 + 6+1 + 7=18$. So $P(D|E)=\frac{n(D\cap E)}{n(E)}=\frac{7}{18}\neq\frac{7}{25}$.

Step5: Calculate $P(F|E)$

$n(F\cap E)=7 + 1=8$, $n(E)=4 + 6+1 + 7=18$. So $P(F|E)=\frac{n(F\cap E)}{n(E)}=\frac{8}{18}=\frac{4}{9}$.

Step6: Calculate $P(E|F)$

$n(E\cap F)=7 + 1=8$, $n(F)=7 + 1+21 + 5=34$. So $P(E|F)=\frac{n(E\cap F)}{n(F)}=\frac{8}{34}=\frac{4}{17}\neq\frac{13}{21}$.

Answer:

$P(E|D)=\frac{7}{25}$, $P(F|E)=\frac{8}{18}$