Answered

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A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at
random. (The first ball is not replaced). Find the probability that both balls are of
the same colour

Sagot :

MrRoyal

Answer:

[tex]P(Same)=\frac{61}{190}[/tex]

Step-by-step explanation:

Given

[tex]Red = 5[/tex]

[tex]White = 6[/tex]

[tex]Black = 9[/tex]

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

[tex]Total = 5 + 6 + 9[/tex]

[tex]Total = 20[/tex]

This is calculated as:

[tex]P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)[/tex]

So, we have:

[tex]P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1} + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}[/tex]

Note that: 1 is subtracted because it is a probability without replacement

[tex]P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1} + \frac{9}{20} * \frac{9- 1}{20- 1}[/tex]

[tex]P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19} + \frac{9}{20} * \frac{8}{19}[/tex]

[tex]P(Same)=\frac{20}{380} + \frac{30}{380} + \frac{72}{380}[/tex]

[tex]P(Same)=\frac{20+30+72}{380}[/tex]

[tex]P(Same)=\frac{122}{380}[/tex]

[tex]P(Same)=\frac{61}{190}[/tex]

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