Question

a box contains 5white and 7 black balls. two successive drawn of 3 balls are made.
1)with replacement
2)without replacement
the probability that the 1st draw would produce white balls and 2nd draw would produce black balls are

Answer 1

Answer on Question #40201, Math, Statistics and Probability

A box contains 5 white and 7 black balls. Two successive drawn of 3 balls are made.

1) with replacement

2) without replacement

The probability that the 1st draw would produce white balls and 2nd draw would produce black balls are

**Solution**

$C_k^n$ - the number of $k$-combinations from a given set of $n$ elements.

$$C_k^n = \frac{n!}{(n - k)! \, k!} = \frac{n(n - 1)(n - 2) \cdots (n - k + 1)}{k!},$$

where $n!$ denotes the factorial of $n$.

**1) When the balls are replaced before 2nd draw**

Total balls in the box = $5 + 7 = 12$.

3 balls can be drawn out of 12 balls in $C_3^{12}$ ways.

3 white balls can be drawn out of 5 white balls is $C_3^5$ ways.

The probability of drawing 3 white balls

$$P(3W) = \frac{C_3^5}{C_3^{12}}.$$

3 balls can be drawn out of 12 balls in $C_3^{12}$ ways.

3 black balls can be drawn out of 7 white balls is $C_3^7$ ways.

The probability of drawing 3 black balls

$$P(3B) = \frac{C_3^7}{C_3^{12}}.$$

Since both the events are dependent, the required probability is

$$P(3W \text{ and } 3B) = \frac{C_3^5}{C_3^{12}} \cdot \frac{C_3^7}{C_3^{12}} = \frac{10}{220} \cdot \frac{35}{220} = 0.007.$$

**2) When the balls are not replaced before 2nd draw**

Total balls in the box = $5 + 7 = 12$.

3 balls can be drawn out of 12 balls in $C_3^{12}$ ways.

3 white balls can be drawn out of 5 white balls is $C_3^5$ ways.

The probability of drawing 3 white balls

$$P(3W) = \frac{C_3^5}{C_3^{12}}.$$

After the first draw, balls left are 9.

3 balls can be drawn out of 9 balls in $C_3^9$ ways.

3 black balls can be drawn out of 7 white balls is $C_3^7$ ways.

The probability of drawing 3 black balls

$$P(3B) = \frac{C_3^7}{C_3^9}.$$

Since both the events are dependent, the required probability is

$$P(3W \text{ and } 3B) = \frac{C_3^5}{C_3^{12}} \cdot \frac{C_3^7}{C_3^9} = \frac{10}{220} \cdot \frac{35}{84} = 0.019.$$