A lens factory produces lenses. The probability of a lens breaking on the first drop is $\frac{1}{2}$. If it doesn't break, the probability of breaking on the second drop is $\frac{7}{10}$. If it still doesn't break, the probability of breaking on the third drop is $\frac{9}{10}$. We want to find the probability that the lens falls three times without breaking.

1. Problem Description

A lens factory produces lenses. The probability of a lens breaking on the first drop is $\frac{1}{2}$. If it doesn't break, the probability of breaking on the second drop is $\frac{7}{10}$. If it still doesn't break, the probability of breaking on the third drop is $\frac{9}{10}$. We want to find the probability that the lens falls three times without breaking.

2. Solution Steps

Let $P(B_i)$ be the probability that the lens breaks on the $i$-th drop, and $P(NB_i)$ be the probability that the lens does not break on the $i$-th drop. We are given:

$P(B_1) = \frac{1}{2}$

$P(B_2 | NB_1) = \frac{7}{10}$

$P(B_3 | NB_1 \cap NB_2) = \frac{9}{10}$

We want to find the probability that the lens doesn't break on the first drop, the second drop, and the third drop. That is $P(NB_1 \cap NB_2 \cap NB_3)$.

First, $P(NB_1) = 1 - P(B_1) = 1 - \frac{1}{2} = \frac{1}{2}$.

Next, $P(NB_2 | NB_1) = 1 - P(B_2 | NB_1) = 1 - \frac{7}{10} = \frac{3}{10}$.

Then, $P(NB_1 \cap NB_2) = P(NB_2 | NB_1) \cdot P(NB_1) = \frac{3}{10} \cdot \frac{1}{2} = \frac{3}{20}$.

Finally, $P(NB_3 | NB_1 \cap NB_2) = 1 - P(B_3 | NB_1 \cap NB_2) = 1 - \frac{9}{10} = \frac{1}{10}$.

Then, $P(NB_1 \cap NB_2 \cap NB_3) = P(NB_3 | NB_1 \cap NB_2) \cdot P(NB_1 \cap NB_2) = \frac{1}{10} \cdot \frac{3}{20} = \frac{3}{200}$.

3. Final Answer

The probability that the lens falls three times without breaking is $\frac{3}{200}$.