Question

use the venn diagram to calculate conditional probabilities. which conditional probabilities are correct? check all that apply. p(d|e) = 6/14, p(e|d) = 7/29, p(d|e) = 7/29, p(e|e) = 8/18, p(e|d) = 11/21

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{n(A\cap B)}{n(B)}$, 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 $P(D|E)$

$n(E)=4 + 0+1 + 7=12$, $n(D\cap E)=0 + 1=1$. So $P(D|E)=\frac{n(D\cap E)}{n(E)}=\frac{1}{12}$.

Step3: Calculate $P(E|D)$

$n(D)=13 + 0+1 + 5=19$, $n(D\cap E)=0 + 1=1$. So $P(E|D)=\frac{n(D\cap E)}{n(D)}=\frac{1}{19}$.

Step4: Calculate $P(D|F)$

$n(F)=21 + 7+1 + 5=34$, $n(D\cap F)=1 + 5=6$. So $P(D|F)=\frac{n(D\cap F)}{n(F)}=\frac{6}{34}=\frac{3}{17}$.

Step5: Calculate $P(F|E)$

$n(E)=4 + 0+1 + 7=12$, $n(F\cap E)=7 + 1=8$. So $P(F|E)=\frac{n(F\cap E)}{n(E)}=\frac{8}{12}=\frac{2}{3}$.

Step6: Calculate $P(E|F)$

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

Let's check the given options one - by - one:

• For $P(D|E)=\frac{1}{12}

eq\frac{6}{14}$, so the first option is incorrect.

• For $P(E|D)=\frac{1}{19}

eq\frac{7}{29}$, so the second option is incorrect.

• For $P(D|E)=\frac{1}{12}

eq\frac{7}{29}$, so the third option is incorrect.

• For $P(F|E)=\frac{8}{12}=\frac{2}{3}=\frac{8}{12}$, this option is correct.
• For $P(E|F)=\frac{8}{34}=\frac{4}{17}

eq\frac{11}{21}$, so the fifth option is incorrect.

Answer:

$P(F|E)=\frac{8}{12}$ is the correct conditional probability among the given options.