Questions: Mark for Review An art gallery has 8 walls showing only paintings by artist A and 4 walls showing only paintings by artist B. Each wall with paintings by artist A has 3 paintings with 4 colors and 2 paintings with fewer than 4 colors, and each wall with paintings by artist B has 2 paintings with 4 colors and 5 paintings with fewer than 4 colors. If a visitor views one of these paintings at random and the painting has fewer than 4 colors, what is the probability that it is by artist A ? (A) 4/17 (B) 4/9 (C) 5/9 (D) 13/17

Solution Steps

First, calculate the total number of paintings with fewer than 4 colors by artist \( A \) and artist \( B \).

• Artist \( A \):
  Each wall has 2 paintings with fewer than 4 colors.
  There are 8 walls.
  Total paintings by artist \( A \) with fewer than 4 colors = \( 8 \times 2 = 16 \).

• Artist \( B \):
  Each wall has 5 paintings with fewer than 4 colors.
  There are 4 walls.
  Total paintings by artist \( B \) with fewer than 4 colors = \( 4 \times 5 = 20 \).

Add the number of paintings with fewer than 4 colors by both artists:

\[ 16 \text{ (by artist \( A \))} + 20 \text{ (by artist \( B \))} = 36 \]

The probability that a randomly selected painting with fewer than 4 colors is by artist \( A \) is given by the ratio of the number of such paintings by artist \( A \) to the total number of such paintings:

\[ P(\text{painting by artist } A \mid \text{fewer than 4 colors}) = \frac{16}{36} = \frac{4}{9} \]

Final Answer