Bag I contains 4 red and 3 black balls and bag II contains 6 red and 5 black balls. One ball is drawn at random from one of the bags, the probability that it is a red ball, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{43}{77} \)
The bag is chosen at random, so each bag is chosen with probability \( \frac{1}{2} \). By the theorem of total probability,
\( \displaystyle P(\text{red})=P(\text{I})\,P(\text{red}|\text{I})+P(\text{II})\,P(\text{red}|\text{II}) \)
\( \displaystyle =\frac{1}{2}\times\frac{4}{7}+\frac{1}{2}\times\frac{6}{11} \)
\( \displaystyle =\frac{2}{7}+\frac{3}{11} \)
\( \displaystyle =\frac{22+21}{77} \)
\( \displaystyle =\frac{43}{77} \)
Shortcut: The total probability is a weighted average of \( \frac{4}{7}\approx 0.571 \) and \( \frac{6}{11}\approx 0.545 \), so it must lie between them. \( \frac{43}{77}\approx 0.558 \) is the only option in that range (\( \frac{34}{77}\approx 0.44 \), \( \frac{35}{68}\approx 0.51 \), \( \frac{33}{68}\approx 0.49 \)).
Tip: Do not pool the balls (10 red out of 18): the bags are chosen with equal probability, not the balls.