Question

Bag A: 4W,3R,2B; Bag B: 2W,4R,3B. One bag chosen randomly, two balls drawn. Probability of one white and one black is

• A. \(\frac{7}{18}\)
• B. \(\frac{2}{9}\)
• C. \(\frac{7}{36}\)
• D. \(\frac{1}{6}\) \bigskip

Hint:

When two bags are involved, always use total probability theorem.

Answer 1

The Correct Option is C

Concept: Use total probability theorem.

Step 1: Bag A probability.
\[ P_A=\frac{\binom{4}{1}\binom{2}{1}}{\binom{9}{2}}=\frac{8}{36}=\frac{2}{9} \]

Step 2: Bag B probability.
\[ P_B=\frac{\binom{2}{1}\binom{3}{1}}{\binom{9}{2}}=\frac{6}{36}=\frac{1}{6} \]

Step 3: Total probability.
\[ \frac12\left(\frac{2}{9}+\frac{1}{6}\right) = \frac12\cdot\frac{7}{18} = \frac{7}{36} \] \[ \boxed{(C)} \]