Question

A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. The probability that first is white and other is black is

• A. $\frac{1}{2}$
• B. 1/3
• C. 1/5
• D. 1/4

Answer 1

Answer: (D)

1/4

Answer 2

Total number of balls in the bag = Number of white balls + Number of black balls = 10 + 15 = 25 balls.

We want to find the probability that the first ball drawn is white and the second ball drawn is black, without replacement.

1. Probability of drawing a white ball first ($P(W_1)$):

There are 10 white balls out of 25 total balls.

$P(W_1) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{10}{25}$

2. Probability of drawing a black ball second, given the first was white and not replaced ($P(B_2 | W_1)$):

After drawing one white ball, there are 24 balls remaining in the bag (25 - 1 = 24).

The number of black balls remains 15.

$P(B_2 | W_1) = \frac{\text{Number of black balls remaining}}{\text{Total number of balls remaining}} = \frac{15}{24}$

3. Probability that the first is white and the other is black ($P(W_1 \text{ and } B_2)$):

This is the product of the probabilities from step 1 and step 2.

$P(W_1 \text{ and } B_2) = P(W_1) \times P(B_2 | W_1)$

$P(W_1 \text{ and } B_2) = \frac{10}{25} \times \frac{15}{24}$

Now, simplify the fractions:

$\frac{10}{25} = \frac{2 \times 5}{5 \times 5} = \frac{2}{5}$

$\frac{15}{24} = \frac{3 \times 5}{3 \times 8} = \frac{5}{8}$

Substitute the simplified fractions back into the probability calculation:

$P(W_1 \text{ and } B_2) = \frac{2}{5} \times \frac{5}{8}$

Multiply the numerators and denominators:

$P(W_1 \text{ and } B_2) = \frac{2 \times 5}{5 \times 8}$

Cancel out the common factor of 5:

$P(W_1 \text{ and } B_2) = \frac{2}{8}$

Simplify the fraction:

$P(W_1 \text{ and } B_2) = \frac{1}{4}$