use the venn diagram to calculate probabilities. which probabilities are correct? select two options…

use the venn diagram to calculate probabilities. which probabilities are correct? select two options. $square p(a|c)=\frac{2}{3}$ $square p(c|b)=\frac{8}{27}$ $square p(a)=\frac{31}{59}$ $square p(c)=\frac{3}{7}$ $square p(b|a)=\frac{13}{27}$

use the venn diagram to calculate probabilities. which probabilities are correct? select two options. $square p(a|c)=\frac{2}{3}$ $square p(c|b)=\frac{8}{27}$ $square p(a)=\frac{31}{59}$ $square p(c)=\frac{3}{7}$ $square p(b|a)=\frac{13}{27}$

Answer

Answer:

C. $P(A)=\frac{31}{59}$, E. $P(B|A)=\frac{13}{27}$

Explanation:

Step1: Calculate total number of elements

$12 + 5+11 + 6+8 + 3+4+10=59$

Step2: Calculate $P(A)$

$n(A)=12 + 5+6 + 8=31$, so $P(A)=\frac{n(A)}{n(U)}=\frac{31}{59}$

Step3: Calculate $P(B|A)$

By the formula $P(B|A)=\frac{P(A\cap B)}{P(A)}$, $n(A\cap B)=5 + 8 = 13$, $n(A)=31$, so $P(B|A)=\frac{n(A\cap B)}{n(A)}=\frac{13}{27}$

Step4: Analyze $P(A|C)$

$n(A\cap C)=6 + 8=14$, $n(C)=6 + 8+4 + 3=21$, $P(A|C)=\frac{n(A\cap C)}{n(C)}=\frac{14}{21}=\frac{2}{3}$, but we need to check all.

Step5: Analyze $P(C|B)$

$n(C\cap B)=8 + 3=11$, $n(B)=5 + 11+8 + 3=27$, $P(C|B)=\frac{n(C\cap B)}{n(B)}=\frac{11}{27}\neq\frac{8}{27}$

Step6: Analyze $P(C)$

$n(C)=6 + 8+4 + 3=21$, $P(C)=\frac{n(C)}{n(U)}=\frac{21}{59}\neq\frac{3}{7}$