Question

List the values of the given random variable X together with the probability distributions.

An urn contains 3 red balls and 7 white balls. Three balls are drawn with replacement. The random variable X is the number of red balls that are drawn.

Answer 1

X follows a binomial distribution

n=3
p=.3

Answer 2

where did the n come from?

Answer 3

and what are the rest of the values?

Answer 4

For me this is a HyperGeomtric distribution because you draw 3 balls but this three balls can come in many combinations posible.

Examples :(red , white , white),(red,red,red)(white , red ,white),etc

the solution is in the file

Any questions you are welcome

Answer 5

it is not hypergeometric...the sampling is done with replacement

Answer 6

it is binomial

Answer 7

"Three balls are drawn with replacement"

three balls...therefore the number of trials n is 3

Answer 8

so they are not dragged randomly , so they hava an order

Answer 9

they are picked randomly and without order

Answer 10

still binomial

Answer 11

"without order" since we only care about the number of red after the 3 trials

Answer 12

can you tell me the definition of . " drawn with replacement "

Answer 13

you pick a ball from the urn...you record the color...you put the ball back into the urn...then you sample again

Answer 14

partMy first language is not english so I must hava misunderstood that e

Answer 15

it the sample was done without replacement then we would have a hypergeometric distribution

Answer 16

sorry my bad

Answer 17

us the binomial distribution

\[P(X=x)={3\choose x}(.3)^x(.7)^{3-x}\]

Answer 18

so what is the answer for P(x=0)

Answer 19

\[P(X=x)={3\choose 0}(.3)^0(.7)^{3}\]

Answer 20

=\[.7^3\]

Answer 21

=.343

Answer 22

Pr(x=0)=(0.7)*(0.7)*(0.7)
Pr(x=3)=(0.3)*(0.3)(0.3)
Pr(x=1)=(0.3)*(0.7)(0.7) *Combinations
Pr(x=2)=(0.3*0.7*0.7)*Combinations

is that ok???

Answer 23

"Pr(x=2)=(0.3*0.7*0.7)*Combinations"

is not correct

Answer 24

Pr(x=0)=(0.7)*(0.7)*(0.7)
Pr(x=3)=(0.3)*(0.3)(0.3)
Pr(x=1)=(0.3)*(0.7)(0.7) *Combinations
Pr(x=2)=(0.3*0.3*0.7)*Combinations

Answer 25

so for x=1 its (3/1)(.3)^1(.7)^3

Answer 26

it is \[3\cdot(.3)^1\cdot(.7)^2\]

Answer 27

@ David correct...what are the combinations though

Answer 28

why .7^2 instead of .7^3?

Answer 29

when x=1 we want 1 read out of 3 that means

1 red and 2 white

Answer 30

it can be in different positions
it can be R W W
W RW
W W R

Answer 31

lol..."red" not 'read'

Answer 32

yes...so 3 ways

Answer 33

\[{3\choose 1}=3\]

Answer 34

\[{3\choose 1}=_3C_1=\frac{3!}{1!(3-1)!}=3\]

Answer 35

so then for x=2 it would be (3/2)(.3)^2(.7)^1

Answer 36

no...
\[P(X=2)={3\choose 2}(.3)^2(.7)^{1}=3\cdot(.3)^2\cdot(.7)^1\]

Answer 37

\[{3\choose 2}\neq3/2\]

Answer 38

Pr(x=0)=(0.7)*(0.7)*(0.7)=0.343
Pr(x=3)=(0.3)*(0.3)(0.3)=0.027
Pr(x=1)=(0.3)*(0.7)(0.7) *Combinations=0.147*3(differen position)=0.441
Pr(x=2)=(0.3*0.3*0.7)*Combinations=0.063**3=0.189

that would be answers i Think

Answer 39

well what does 3 mean?
2

Answer 40

@David..you are correct

Answer 41

they can come in 3 diferrent ways
RWW
WWR
WRW

Answer 42

\[{3\choose 2}=_3C_2=\frac{3!}{2!(3-2)!}=3\]

Answer 43

\[{3\choose 2}=_3C_2=\frac{3!}{2!(3-2)!}=\frac{3!}{2!\cdot 1!}=\frac{3\cdot2\cdot1}{2\cdot1\cdot1}=3\]

Answer 44

when n and x are small you can list them as David did

Answer 45

so how would i set up x=3?

Answer 46

x=3 is RRR

only one way to do that so ...

\[1\cdot (.3)^3(.7)^0=.3^3\]

Answer 47

or

\[P(X=3)={3\choose 3}(.3)^3(.7)^0=1\cdot(.3)^3\cdot(.7)^0\]

Answer 48

.027?

Answer 49

yes

Answer 50

Remember one way to verify if u are correct is summing up the probabilities then you have to get ONE "1" otherwise something is wrong