Question

practice analyzing two - way tables to determine condition probabilities and independent events. consider the two - way table below. find p(b|c). c d total a 15 21 36 b 9 25 34 total 24 46 70 p(b|c)=\frac{9}{34}=0.26 p(b|c)=\frac{9}{24}=0.38 p(b|c)=\frac{9}{70}=0.13 p(b|c)=\frac{9}{58}=0.16

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(B|C)=\frac{P(B\cap C)}{P(C)}$. In the context of a two - way table, $P(B|C)=\frac{n(B\cap C)}{n(C)}$, where $n(B\cap C)$ is the number of elements in both $B$ and $C$, and $n(C)$ is the number of elements in $C$.

Step2: Identify values from the table

From the two - way table, $n(B\cap C) = 9$ (the value in the cell where row $B$ and column $C$ intersect) and $n(C)=24$ (the total number of elements in column $C$).

Step3: Calculate the conditional probability

$P(B|C)=\frac{9}{24}=0.375\approx0.38$.

Answer:

$P(B|C)=\frac{9}{24}=0.38$