19) In a box there are four bags: the first bag contains 4 white balls and 3 black balls, the second 2 white balls and 4 black balls, the third 6 white balls and 9 black balls, the fourth 5 white balls and 10 black balls. A bag is drawn at random, and from this, a ball is drawn at random. Knowing that a white ball has been drawn, which bag is it most likely that was chosen?
(A) The first
(B) the second
(C) the third
(D) the fourth

(E) all bags have the same probability of being chosen.
19) The answer is (A).

![](https://cdn.mathpix.com/cropped/2024_04_17_23f072eac9b2d02bc457g-3.jpg?height=368&width=488&top_left_y=770&top_left_x=1435)

Let $p_{1}, p_{2}, p_{3}, p_{4}$ be the probability of drawing a white ball from the first, second, third, and fourth bag, respectively. We have

$$
p_{1}=\frac{1}{4} \cdot \frac{4}{7} \quad p_{2}=\frac{1}{4} \cdot \frac{2}{6} \quad p_{3}=\frac{1}{4} \cdot \frac{6}{15} \quad p_{4}=\frac{1}{4} \cdot \frac{5}{15} .
$$

Since it is known that a white ball was drawn, the probability that it was drawn from the i-th bag is $\frac{p_{i}}{p_{1}+p_{2}+p_{3}+p_{4}}$. The bag from which the ball is most likely to have been drawn is thus the first; this bag is the one in which the ratio of the number of white balls to the number of black balls is the highest.