Question

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

Explanation:

Step1: Recall the formula for conditional probability

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

Step2: Calculate \(P(D|F)\)

First, find \(n(D\cap F) = 5 + 1=6\) and \(n(F)=21 + 7+1 + 5=34\). Then \(P(D|F)=\frac{n(D\cap F)}{n(F)}=\frac{6}{34}\).

Step3: Calculate \(P(E|D)\)

Find \(n(E\cap D)=6 + 1=7\) and \(n(D)=13+6 + 1+5=25\). Then \(P(E|D)=\frac{n(E\cap D)}{n(D)}=\frac{7}{25}\).

Step4: Calculate \(P(D|E)\)

Find \(n(D\cap E)=6 + 1=7\) and \(n(E)=4 + 6+1 + 7=18\). Then \(P(D|E)=\frac{n(D\cap E)}{n(E)}=\frac{7}{18}
eq\frac{7}{25}\).

Step5: Calculate \(P(F|E)\)

Find \(n(F\cap E)=7 + 1=8\) and \(n(E)=4 + 6+1 + 7=18\). Then \(P(F|E)=\frac{n(F\cap E)}{n(E)}=\frac{8}{18}\).

Step6: Calculate \(P(E|F)\)

Find \(n(E\cap F)=7 + 1=8\) and \(n(F)=21 + 7+1 + 5=34\). Then \(P(E|F)=\frac{n(E\cap F)}{n(F)}=\frac{8}{34}
eq\frac{13}{21}\).

Answer:

\(P(D|F)=\frac{6}{34}\), \(P(E|D)=\frac{7}{25}\), \(P(F|E)=\frac{8}{18}\)