Question

practice analyzing two-way tables to determine conditional probabilities and independent events.

consider the two-way table below.

\

$$\begin{tabular}{|c|c|c|c|} \\hline & c & d & total \\\\ \\hline a & 15 & 21 & 36 \\\\ \\hline b & 9 & 25 & 34 \\\\ \\hline total & 24 & 46 & 70 \\\\ \\hline \\end{tabular}$$

find \\(p(b|c)\\).

\\(p(b|c) = \frac{9}{34} \approx 0.26\\)

\\(p(b|c) = \frac{9}{24} \approx 0.38\\)

\\(p(b|c) = \frac{9}{70} \approx 0.13\\)

\\(p(b|c) = \frac{9}{58} \approx 0.16\\)

Explanation:

Identify the conditional probability formula

Using the Conditional Probability Calculation and Two-Way Frequency Tables knowledge points

$$ P(B|C) = \frac{P(B \cap C)}{P(C)} = \frac{n(B \cap C)}{n(C)} $$

Extract values from the two-way table

Using the Two-Way Frequency Tables knowledge point

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Calculate the conditional probability

Using the Conditional Probability Calculation knowledge point

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

Answer:

• P(B|C) = \(\frac{9}{34} = 0.26\)
• P(B|C) = \(\frac{9}{24} = 0.38\) (Correct answer)
• P(B|C) = \(\frac{9}{70} = 0.13\)
• P(B|C) = \(\frac{9}{58} = 0.16\)