In a survey of 1438 multigenerational households, the skip-generation column contains 53 White, 5 Hispanic, 34 Black and 7 Asian families, for a column total of 99. What is the probability that a randomly selected multigenerational family is White, given that it is a skip-generation family? Round to three decimal places.

Step-by-step solution

1. 1

   Identify which variable is the condition. The words "given that it is a skip-generation family" put skip generation after the bar, so you want

   $P(\text{White}\mid\text{Skip})=\frac{P(\text{White and Skip})}{P(\text{Skip})}$

   This time the condition is a column, not a row, so the column total becomes the denominator.

2. 2

   Read the column total. Summing the skip-generation column: $53+5+34+7=99$. That matches the printed total, so $n(\text{Skip})=99$.

3. 3

   Read the joint cell. The White row meets the skip-generation column at $53$, so $n(\text{White and Skip})=53$.

4. 4

   Divide with counts instead of proportions. Because both the numerator and the denominator would be divided by the same $1438$, the grand total cancels and you can work directly with counts:

   $P(\text{White}\mid\text{Skip})=\frac{53/1438}{99/1438}=\frac{53}{99}=0.5353\ldots\approx 0.535$

   $\boxed{0.535}$

5. 5

   Note that the reverse conditional is different. $P(\text{Skip}\mid\text{White})=\tfrac{53}{786}\approx 0.067$, nowhere near $0.535$. Swapping the two events in a conditional probability changes the answer, so always check which count belongs on the bottom.