Assume a jar has five red marbles and three black marbles. Draw out two marbles with and without replacement. Find the requested probabilities. (Enter the probabilities as fractions.) (a) P(two red marbles) with replacement without replacement (b) P(two black marbles) with replacement without replacement (c) P(one red and one black marble) with replacement without replacement (d) P(red on the first draw and black on the second draw) with replacement without replacement

Question

12-03-2021
Mathematics

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Assume a jar has five red marbles and three black marbles. Draw out two marbles with and without replacement. Find the requested probabilities. (Enter the probabilities as fractions.) (a) P(two red marbles) with replacement without replacement (b) P(two black marbles) with replacement without replacement (c) P(one red and one black marble) with replacement without replacement (d) P(red on the first draw and black on the second draw) with replacement without replacement

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MrRoyal MrRoyal · Answer 1 · 2021-03-15T03:16:19+01:00

Answer:

For probabilities with replacement

[tex]P(2\ Red) = \frac{25}{64}[/tex]

[tex]P(2\ Black) = \frac{9}{64}[/tex]

[tex]P(1\ Red\ and\ 1\ Black) = \frac{15}{32}[/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}[/tex]

For probabilities without replacement

[tex]P(2\ Red) = \frac{5}{14}[/tex]

[tex]P(2\ Black) = \frac{3}{28}[/tex]

[tex]P(1\ Red\ and\ 1\ Black) = \frac{15}{28}[/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}[/tex]

Step-by-step explanation:

Given

[tex]Marbles = 8[/tex]

[tex]Red = 5[/tex]

[tex]Black = 3[/tex]

For probabilities with replacement

(a) P(2 Red)

This is calculated as:

[tex]P(2\ Red) = P(Red)\ and\ P(Red)[/tex]

[tex]P(2\ Red) = P(Red)\ *\ P(Red)[/tex]

So, we have:

[tex]P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)}{Total}\\[/tex]

[tex]P(2\ Red) = \frac{5}{8} * \frac{5}{8}[/tex]

[tex]P(2\ Red) = \frac{25}{64}[/tex]

(b) P(2 Black)

This is calculated as:

[tex]P(2\ Black) = P(Black)\ and\ P(Black)[/tex]

[tex]P(2\ Black) = P(Black)\ *\ P(Black)[/tex]

So, we have:

[tex]P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)}{Total}[/tex]

[tex]P(2\ Black) = \frac{3}{8}\ *\ \frac{3}{8}[/tex]

[tex]P(2\ Black) = \frac{9}{64}[/tex]

(c) P(1 Red and 1 Black)

This is calculated as:

[tex]P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)][/tex]

[tex]P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)][/tex]

[tex]P(1\ Red\ and\ 1\ Black) = 2[P(Red)\ *\ P(Black)][/tex]

So, we have:

[tex]P(1\ Red\ and\ 1\ Black) = 2*[\frac{5}{8} *\frac{3}{8}][/tex]

[tex]P(1\ Red\ and\ 1\ Black) = 2*\frac{15}{64}[/tex]

[tex]P(1\ Red\ and\ 1\ Black) = \frac{15}{32}[/tex]

(d) P(1st Red and 2nd Black)

This is calculated as:

[tex]P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)][/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)[/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total}[/tex]

So, we have:

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{8}[/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}[/tex]

For probabilities without replacement

(a) P(2 Red)

This is calculated as:

[tex]P(2\ Red) = P(Red)\ and\ P(Red)[/tex]

[tex]P(2\ Red) = P(Red)\ *\ P(Red)[/tex]

So, we have:

[tex]P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)-1}{Total-1}[/tex]

We subtracted 1 because the number of red balls (and the total) decreased by 1 after the first red ball is picked.

[tex]P(2\ Red) = \frac{5}{8} * \frac{4}{7}[/tex]

[tex]P(2\ Red) = \frac{5}{2} * \frac{1}{7}[/tex]

[tex]P(2\ Red) = \frac{5}{14}[/tex]

(b) P(2 Black)

This is calculated as:

[tex]P(2\ Black) = P(Black)\ and\ P(Black)[/tex]

[tex]P(2\ Black) = P(Black)\ *\ P(Black)[/tex]

So, we have:

[tex]P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)-1}{Total-1}[/tex]

We subtracted 1 because the number of black balls (and the total) decreased by 1 after the first black ball is picked.

[tex]P(2\ Black) = \frac{3}{8}\ *\ \frac{2}{7}[/tex]

[tex]P(2\ Black) = \frac{3}{4}\ *\ \frac{1}{7}[/tex]

[tex]P(2\ Black) = \frac{3}{28}[/tex]

(c) P(1 Red and 1 Black)

This is calculated as:

[tex]P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)][/tex]

[tex]P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)][/tex]

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

So, we have:

[tex]P(1\ Red\ and\ 1\ Black) = [\frac{5}{8} *\frac{3}{7}] + [\frac{3}{8} *\frac{5}{7}][/tex]

[tex]P(1\ Red\ and\ 1\ Black) = [\frac{15}{56} ] + [\frac{15}{56}][/tex]

[tex]P(1\ Red\ and\ 1\ Black) = \frac{30}{56}[/tex]

[tex]P(1\ Red\ and\ 1\ Black) = \frac{15}{28}[/tex]

(d) P(1st Red and 2nd Black)

This is calculated as:

[tex]P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)][/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)[/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total-1}[/tex]

So, we have:

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{7}[/tex]

[tex]P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}[/tex]