Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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]

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.