We have a box with 4 products, 3 first-grade and 1 second-grade. Two products are drawn sequentially without replacement. We want to find the conditional probability $P(B|A)$, where $A$ is the event that the first drawn product is first-grade and $B$ is the event that the second drawn product is first-grade.

1. Problem Description

We have a box with 4 products, 3 first-grade and 1 second-grade. Two products are drawn sequentially without replacement. We want to find the conditional probability $P(B|A)$, where $A$ is the event that the first drawn product is first-grade and $B$ is the event that the second drawn product is first-grade.

2. Solution Steps

We want to calculate $P(B|A)$, which is the probability that the second product drawn is first-grade, given that the first product drawn was first-grade.

By definition of conditional probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

First, let's find $P(A)$. The probability that the first product drawn is first-grade is the number of first-grade products divided by the total number of products:

$P(A) = \frac{3}{4}$

Next, let's find $P(A \cap B)$. This is the probability that both the first and second products drawn are first-grade.

We can calculate this as follows:

$P(A \cap B) = P(A) \cdot P(B|A)$

However, it can also be viewed as the probability of picking a first grade product first and then a first grade product second.

$P(A \cap B) = \frac{\text{Number of ways to pick two first-grade products}}{\text{Number of ways to pick two products}}$

The number of ways to pick two first-grade products is $3/4 \times 2/3 = 6/12 = 1/2$. We have three choices for the first product and then two choices for the second first grade product out of a total of four products initially and then three products after the first one is taken.

$P(A \cap B) = \frac{3}{4} \cdot \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$

Now, we can compute $P(B|A)$:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1/2}{3/4} = \frac{1}{2} \cdot \frac{4}{3} = \frac{4}{6} = \frac{2}{3}$

Alternatively, given that the first product drawn was first-grade, there are now 3 products remaining in the box, 2 of which are first-grade and 1 of which is second-grade. Therefore, the probability that the second product drawn is first-grade is $\frac{2}{3}$.

3. Final Answer

The conditional probability $P(B|A)$ is $\frac{2}{3}$.