Let's solve the two parts of this question step by step.

Problem Statement

We have:

• 3 black balls (B)
• 4 white balls (W)

Two balls are drawn without replacement, and we need to find:

1. The probability that the second ball drawn is white.
2. The conditional probability that the first ball drawn is white given that the second ball is white.

Part (1): Probability that the second ball is white

To calculate this, we need to consider all possible ways the second ball could be white.

Case 1: First ball is black, and second ball is white

• Probability of drawing a black ball first = $\frac{3}{7}$
• Probability of drawing a white ball second (after a black ball) = $\frac{4}{6}$

So, the probability for this case is: $P(\text{Black first, White second}) = \frac{3}{7} \times \frac{4}{6}$

Case 2: First ball is white, and second ball is white

• Probability of drawing a white ball first = $\frac{4}{7}$
• Probability of drawing another white ball second (after a white ball) = $\frac{3}{6}$

So, the probability for this case is: $P(\text{White first, White second}) = \frac{4}{7} \times \frac{3}{6}$

Adding both cases gives the total probability that the second ball is white: $P(\text{Second ball is white}) = \frac{3}{7} \times \frac{4}{6} + \frac{4}{7} \times \frac{3}{6}$

$P(\text{Second ball is white}) = \frac{12}{42} + \frac{12}{42} = \frac{24}{42} = \frac{4}{7}$

Part (2): Conditional probability that the first ball is white, given that the second ball is white

We use the conditional probability formula: $P(\text{First is white} \mid \text{Second is white}) = \frac{P(\text{First is white and Second is white})}{P(\text{Second is white})}$

• $P(\text{First is white and Second is white})$: This is just the second case calculated earlier: $P(\text{First is white and Second is white}) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42}$

• $P(\text{Second is white})$: We already calculated this as $\frac{4}{7}$.

Substitute these into the formula: $P(\text{First is white} \mid \text{Second is white}) = \frac{\frac{12}{42}}{\frac{4}{7}} = \frac{12}{42} \times \frac{7}{4} = \frac{12}{24} = \frac{1}{2}$

Final Answers

1. The probability that the second ball is white: $\frac{4}{7}$.
2. The conditional probability that the first ball is white given the second ball is white: $\frac{1}{2}$.

Do you want further explanation or a breakdown of any step?

Related Questions:

1. What is the probability of both balls being black?
2. How does the problem change if the balls are drawn with replacement?
3. Can we find the probability of drawing one ball of each color in any order?
4. How does the conditional probability change if the total number of balls increases?
5. What happens to the probabilities if there are an equal number of black and white balls?

Tip: For conditional probabilities, always focus on the "given" part to narrow down the outcomes!