Question

use the contingency table below to find the probability of the following events.
a. a|b
b. a|b
c. a|b
d. are events a and b independent?
a. p(a|b) = 0.33 (round to two decimal places as needed.)
b. p(a|b) = \boxed{} (round to two decimal places as needed.)
contingency table: rows b, b; columns a, a. b row: a=10, a=20; b row: a=20, a=40

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(X|Y) = \frac{P(X \cap Y)}{P(Y)} \). For \( P(A|B') \), we need \( n(A \cap B') \) and \( n(B') \) (using counts from the contingency table, since \( P(X \cap Y)=\frac{n(X \cap Y)}{N} \) and \( P(Y)=\frac{n(Y)}{N} \), the \( N \) cancels out, so we can use counts directly: \( P(A|B') = \frac{n(A \cap B')}{n(B')} \)).

Step2: Identify Counts from Table

From the contingency table:

• \( n(A \cap B') = 20 \) (A and B' intersection)
• \( n(B') = n(A \cap B') + n(A' \cap B') = 20 + 40 = 60 \) (total in B' row)

Step3: Calculate Probability

Substitute into the formula: \( P(A|B') = \frac{20}{60} \approx 0.33 \) (Wait, no, wait: Wait, the table: Let's recheck. Wait, the rows are B and B', columns A and A'. So for B': A is 20, A' is 40. So \( n(B') = 20 + 40 = 60 \), \( n(A \cap B') = 20 \). So \( \frac{20}{60} = \frac{1}{3} \approx 0.33 \)? Wait, but maybe I misread. Wait, the user's table: Let's parse the table. The columns are A and A', rows are B and B'. So:

• B row: A=10, A'=20
• B' row: A=20, A'=40

So \( n(B') = 20 + 40 = 60 \), \( n(A \cap B') = 20 \). So \( P(A|B') = \frac{20}{60} = \frac{1}{3} \approx 0.33 \)? Wait, but the first part (a) was \( P(A|B) = \frac{10}{10 + 20} = \frac{10}{30} \approx 0.33 \). For (b), \( P(A|B') = \frac{20}{20 + 40} = \frac{20}{60} \approx 0.33 \)? Wait, no, \( 20/60 = 0.333... \), rounded to two decimal places is 0.33? Wait, 20 divided by 60 is 0.333..., so two decimal places is 0.33. Wait, but maybe I made a mistake. Wait, let's recalculate: 20 ÷ 60 = 0.3333..., so rounded to two decimal places is 0.33.

Answer:

0.33