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) - QAmmunity.org

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)

asked Sep 24, 2022 179k views

1 Answer

Answer:

For probabilities with replacement

$P(2\ Red) = (25)/(64)$

$P(2\ Black) = (9)/(64)$

$P(1\ Red\ and\ 1\ Black) = (15)/(32)$

$P(1st\ Red\ and\ 2nd\ Black) = (15)/(64)$

For probabilities without replacement

$P(2\ Red) = (5)/(14)$

$P(2\ Black) = (3)/(28)$

$P(1\ Red\ and\ 1\ Black) = (15)/(28)$

$P(1st\ Red\ and\ 2nd\ Black) = (15)/(56)$

Explanation:

Given

Marbles = 8

Red = 5

Black = 3

For probabilities with replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = (n(Red))/(Total) \ *\ (n(Red))/(Total)\\$

$P(2\ Red) = (5)/(8) * (5)/(8)$

$P(2\ Red) = (25)/(64)$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = (n(Black))/(Total)\ *\ (n(Black))/(Total)$

$P(2\ Black) = (3)/(8)\ *\ (3)/(8)$

$P(2\ Black) = (9)/(64)$

(c) P(1 Red and 1 Black)

This is calculated as:

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

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

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

So, we have:

$P(1\ Red\ and\ 1\ Black) = 2*[(5)/(8) *(3)/(8)]$

$P(1\ Red\ and\ 1\ Black) = 2*(15)/(64)$

$P(1\ Red\ and\ 1\ Black) = (15)/(32)$

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

This is calculated as:

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

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = (n(Red))/(Total) *\ (n(Black))/(Total)$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = (5)/(8) *(3)/(8)$

$P(1st\ Red\ and\ 2nd\ Black) = (15)/(64)$

For probabilities without replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = (n(Red))/(Total) \ *\ (n(Red)-1)/(Total-1)$

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

$P(2\ Red) = (5)/(8) * (4)/(7)$

$P(2\ Red) = (5)/(2) * (1)/(7)$

$P(2\ Red) = (5)/(14)$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = (n(Black))/(Total)\ *\ (n(Black)-1)/(Total-1)$

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

$P(2\ Black) = (3)/(8)\ *\ (2)/(7)$

$P(2\ Black) = (3)/(4)\ *\ (1)/(7)$

$P(2\ Black) = (3)/(28)$

(c) P(1 Red and 1 Black)

This is calculated as:

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

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

$P(1\ Red\ and\ 1\ Black) = [(n(Red))/(Total)\ *\ (n(Black))/(Total-1)]\ +\ [(n(Black))/(Total)\ *\ (n(Red))/(Total-1)]$

So, we have:

$P(1\ Red\ and\ 1\ Black) = [(5)/(8) *(3)/(7)] + [(3)/(8) *(5)/(7)]$

$P(1\ Red\ and\ 1\ Black) = [(15)/(56) ] + [(15)/(56)]$

$P(1\ Red\ and\ 1\ Black) = (30)/(56)$

$P(1\ Red\ and\ 1\ Black) = (15)/(28)$

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

This is calculated as:

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

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = (n(Red))/(Total) *\ (n(Black))/(Total-1)$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = (5)/(8) *(3)/(7)$

$P(1st\ Red\ and\ 2nd\ Black) = (15)/(56)$

Smakus answered Sep 29, 2022

by Smakus

7.9k points

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