Question

assess2/?cid=59241&aid=4304153#/fullQuestion #6 (16 points - 2 each)A box contains 7 red balls, 4 green balls, and 2 blue balls. You pull 2 balls from the box (one at atime) WITHOUT replacement.**LEAVE ALL ANSWERS AS FRACTIONS**Find the probability of the following:a.) P(Red on 1st ball AND Red on 2nd ball) =b.) P(Green on 1st ball AND Red on 2nd ball) =c.) P(Blue on 1st ball AND Green on 2nd ball) =d.) What is the probability of drawing 2 green balls in your 2 pulls?e.) What is the probability of selecting a red ball on your second pull, given a red ball was alreadyselected on the first pull?f.) What is the probability of drawing one red ball and one green ball (in either order)?g.) What is the probability or selecting two balls of the same color?h.) What is the probability or selecting two balls of different colors?

Answer 1

we know that

A box contains 7 red balls, 4 green balls, and 2 blue balls

You pull 2 balls from the box (one at a time) WITHOUT replacement

so

Part a

P(Red on 1st ball AND Red on 2nd ball)

step 1

total balls=7+4+2=13 balls

red balls=7

P(Red on 1st ball)=7/13

step 2

total balls=6+4+2=12 balls

red bals=6

P( Red on 2nd ball)=6/12=1/2

so

P(Red on 1st ball AND Red on 2nd ball)=(7/13)*(1/2)=7/26

Part b

P(Green on 1st ball AND Red on 2nd ball) =

step 1

total balls=7+4+2=13 balls

green balls=4

P(Green on 1st ball)=4/13

step 2

total balls=7+3+2=12

red balls=7

P(Red on 2nd ball)=7/12

so

P(Green on 1st ball AND Red on 2nd ball) =(4/13)*(7/12)=7/39

Part c

P(Blue on 1st ball AND Green on 2nd ball) =

step 1

total balls=7+4+2=13 balls

Blue balls=2

P(Blue on 1st ball)=2/13

step 2

total balls=7+4+1=12

green balls=4

P(Green on 2nd ball)=4/12=1/3

so

P(Blue on 1st ball AND Green on 2nd ball) =(2/13)*(1/3)=2/39

Part d

What is the probability of drawing 2 green balls in your 2 pulls?

step 1

total balls=13

green balls=4

P(Green on 1st ball)=4/13

P(Green on 2ns ball)=3/12

so

probability of drawing 2 green balls in your 2 pulls is

P=(4/13)*(3/12)=1/13

Part e

What is the probability of selecting a red ball on your second pull, given that a red ball was already selected on the first pull?

second pull

total balls=12

red balls=6

P=6/12=1/2

Part f

What is the probability of drawing one red ball and one green ball (in either order)?

First case

one red ball-one green ball

P=(7/13)*(4/12)=(7/13)*(1/3)=7/39

Second case

one green ball-one red ball

P=(4/13)*(7/12)=7/39

therefore

the probability is

P=(7/39)+(7/39)=14/39

Part g

What is the probability of selecting two balls of the same color?

step 1

Find out the total outcomes

Find out 13C2

$13C2=\frac{13!}{2!(13-2)!}=78$

step 2

Find out 7C2 (red balls)

$7C2=\frac{7!}{2!(7-2)!}=21$

step 3

Find out 4C2 (green balls)

$4C2=\frac{4!}{2!(4-2)!}=6$

step 4

Find out 2C2 (blue balls)

$2C2=\frac{2!}{2!(2-2)!}=1$

therefore

the probability is equal to

P=(21+6+1)/78

P=28/78

simplify

P=14/39

Part h

What is the probability of selecting two balls of different colors?

we know that

The probability of selecting two balls of the same color is 14/39

therefore

the probability of selecting two balls of different colors is

P=1-14/39

P=25/39