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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

Sagot :

MrRoyal

Answer:

[tex]Probability = 0.4[/tex]

Step-by-step explanation:

Given

[tex]White = 6[/tex]

[tex]Black = 4[/tex]

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:

[tex]P(Black\ Only) = \frac{Black}{Total} * \frac{Black - 1}{Total - 1}[/tex]

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

[tex]P(Black\ Only) = \frac{4}{10} * \frac{4- 1}{10 - 1}[/tex]

[tex]P(Black\ Only) = \frac{4}{10} * \frac{3}{9}[/tex]

[tex]P(Black\ Only) = \frac{2}{5} * \frac{1}{3}[/tex]

[tex]P(Black\ Only) = \frac{2}{15}[/tex]

The probability of White and Black is:

[tex]P(Black\ and\ White) = \frac{Black}{Total} * \frac{White}{Total - 1}[/tex]

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

[tex]P(Black\ and\ White) = \frac{4}{10} * \frac{6}{10- 1}[/tex]

[tex]P(Black\ and\ White) = \frac{4}{10} * \frac{6}{9}[/tex]

[tex]P(Black\ and\ White) = \frac{2}{5} * \frac{2}{3}[/tex]

[tex]P(Black\ and\ White) = \frac{4}{15}[/tex]

So, the required probability is:

[tex]Probability = P(Black\ Only) + P(Black\ and\ White)[/tex]

[tex]Probability = \frac{2}{15} + \frac{4}{15}[/tex]

[tex]Probability = \frac{2+4}{15}[/tex]

[tex]Probability = \frac{6}{15}[/tex]

[tex]Probability = 0.4[/tex]

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