Question

practice analyzing two - way tables to determine conditional probabilities and independent events. consider the two - way table below. find ( p(b|c) ). ( 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 the formula for conditional probability

The formula for conditional probability is \( P(B|C)=\frac{n(B\cap C)}{n(C)} \), 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 the values from the table

From the table, \( n(B\cap C) = 9 \) (the intersection of row \( B \) and column \( C \)) and \( n(C)=24 \) (the total of column \( C \)).

Step3: Calculate the probability

Substitute the values into the formula: \( P(B|C)=\frac{9}{24} = 0.375\approx0.38 \)

Answer:

\( P(B|C)=\frac{9}{24} = 0.38 \) (the second option)