A survey of 1438 multigenerational households is broken down by race and by household type. The Black row shows 114 two-adult-generation, 34 skip-generation, and 102 three-or-more-generation families, for a row total of 250. What is the probability that a randomly selected Black multigenerational family is a two-adult-generation family? Round to three decimal places.

Step-by-step solution

1. 1

   Read what the conditioning word does. The phrase "a randomly selected Black multigenerational family" tells you the selection is already known to come from the Black row. That is a conditional probability $P(\text{2 adult gens}\mid\text{Black})$, and the single most common mistake here is dividing by the grand total $1438$ instead of by the row total.

2. 2

   Shrink the sample space to the given row. Once you know the family is Black, the only outcomes still possible are the $250$ Black households. This row total is the new denominator:

   $P(A\mid B)=\frac{n(A\text{ and }B)}{n(B)}$

3. 3

   Count the favourable outcomes inside that row. The cell where the Black row meets the "2 Adult Gens" column holds $114$ households, so $n(A\text{ and }B)=114$.

4. 4

   Divide and round.

   $P(\text{2 adult gens}\mid\text{Black})=\frac{114}{250}=0.456$

   The division is exact, so no rounding error is introduced: $\tfrac{114}{250}=\tfrac{57}{125}$.

   $\boxed{0.456}$

5. 5

   Sanity-check against the unconditional probability. If you had wrongly used the grand total you would get $114/1438\approx 0.079$, a very different number. Also note the three conditional probabilities down the Black row must sum to $1$: $\tfrac{114}{250}+\tfrac{34}{250}+\tfrac{102}{250}=\tfrac{250}{250}=1$, which confirms the row total was read correctly.