Question

In a bag, there are six white balls and four black balls. I take out two balls at random, one at a time, without replacement. What is the probability that the second ball I take is black

Answer 1

Answer:

$Probability = 0.4$

Step-by-step explanation:

Given

$White = 6$

$Black = 4$

Required

Probability that second is black

This selection can be represented as:

(Black and Black) or (White and Black)

The probability of Black and Black is:

$P(Black\ Only) = \frac{Black}{Total} * \frac{Black - 1}{Total - 1}$

1 is subtracted from the second fraction because it is probability without replacement

$P(Black\ Only) = \frac{4}{10} * \frac{4- 1}{10 - 1}$

$P(Black\ Only) = \frac{4}{10} * \frac{3}{9}$

$P(Black\ Only) = \frac{2}{5} * \frac{1}{3}$

$P(Black\ Only) = \frac{2}{15}$

The probability of White and Black is:

$P(Black\ and\ White) = \frac{Black}{Total} * \frac{White}{Total - 1}$

1 is subtracted from the second fraction because it is probability without replacement

$P(Black\ and\ White) = \frac{4}{10} * \frac{6}{10- 1}$

$P(Black\ and\ White) = \frac{4}{10} * \frac{6}{9}$

$P(Black\ and\ White) = \frac{2}{5} * \frac{2}{3}$

$P(Black\ and\ White) = \frac{4}{15}$

So, the required probability is:

$Probability = P(Black\ Only) + P(Black\ and\ White)$

$Probability = \frac{2}{15} + \frac{4}{15}$

$Probability = \frac{2+4}{15}$

$Probability = \frac{6}{15}$

$Probability = 0.4$