Question

help

Answer 1

@freckles

Answer 2

you want to find P ( X >= 4)

Answer 3

how can I do that

Answer 4

P( X>=4) = P( X = 4) + P( X = 5) + P( X= 6) + P( X = 7 ) + P( X = 8)

Answer 5

you can approach this as a binomial distribution problem

Answer 6

it would be easier to use the complement, since there are less terms to add.
P( X >= 4) = 1 - P( X <= 3)

Answer 7

does your teacher allow you to use a calculator?

Answer 8

yes

Answer 9

the calculator has a built in function which allows you to find the cumulative binomial probability

Answer 10

is it the x! ? because I believe he showed us something with that function

Answer 11

huh?

Answer 12

What kind of calculator do you have?

Answer 13

scientific

Answer 14

do you have a TI calculator (like a TI83, 84)?

Answer 15

TI-30XA

Answer 16

ok I'm not really familiar with that calculator. Does it have a stats menu or anything?

Answer 17

no idea :/

Answer 18

it doesn't look it

Answer 19

are you allowed to use online calculators?

Answer 20

not for the final, but I can now since it's just review

Answer 21

so it's probably best to practice with the calculator you have in your hands

Answer 22

i'm still not really sure how to do the problem though

Answer 23

as perl is saying, you need to calculate the following probabilities

P(X = 0)
P(X = 1)
P(X = 2)
P(X = 3)

to calculate the probability P(X = k) we use this formula

(n C k)*(p)^k*(1-p)^(n-k)

the n C k refers to the combination formula shown below

\[\Large _n C _k = \frac{n!}{k!(n-k)!}\]
so this is where the factorial will factor in

Answer 24

how do I know what n and k are?

Answer 25

so to calculate P(X = 0), the value of k will be zero
n = 8 always since there are 8 questions
p = 1/5 = 0.2 because there is 1 in 5 chance of getting it right

Answer 26

for P(X = 1) through P(X = 3) the value of n and p will remain the same. Only k will change

Answer 27

for p(x=0) I got 1

Answer 28

what is (n C k) equal to when n = 8 and k = 0?

Answer 29

oh I'm just realizing there's a "nCr" label above the "8" key

see this page


https://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=74599

Answer 30

yes i see it

Answer 31

so you can type

8, 2ND, 8, 0, enter

and out will come "1". That is not the value of P(X = 0). That's just part of it

Answer 32

so i've got 1, 8,28,56

Answer 33

oh :/

Answer 34

remember: all probabilities are between 0 and 1 (inclusive)

Answer 35

you have to use this formula

(n C k)*(p)^k*(1-p)^(n-k)

Answer 36

the n C k part is simply a number found after using the n C r function

Answer 37

n = 8
p = 0.2
k = varies from 0 to 3

Answer 38

oh lord

Answer 39

now I'm lost

Answer 40

I'm going to calculate P(X = 0)

before you found that n C k = 1 when n = 8 and k = 0

so

(n C k)*(p)^k*(1-p)^(n-k)

(8 C 0)*(0.2)^0*(1-0.2)^(8 - 0)

(1)*(0.2)^0*(1-0.2)^(8 - 0)

(1)*(0.2)^0*(0.8)^(8)

(1)*1*0.16777216

0.16777216

Answer 41

P(X = 0) = 0.16777216

Answer 42

keep n and p the same (n = 8, p = 0.2) and plug in k = 1 this time

use the same formula

(n C k)*(p)^k*(1-p)^(n-k)

and tell me what P(X = 1) is equal to

Answer 43

I really have no idea, I'm sorry, thank you for trying though :/

Answer 44

you're seeing how n C k = 1 when n = 8 and k = 0 right?

Answer 45

yes, you just plug them into the equation

Answer 46

or you can use the nCr button on your calculator

Answer 47

yep

Answer 48

so we replace the n C k portion with just '1'

\[\Large \left(_n C _k\right)*(p)^k*(1-p)^{n-k}\]

\[\Large \left({\color{red}{_n C _k}}\right)*(p)^k*(1-p)^{n-k}\]

\[\Large \left({\color{red}{_8 C _0}}\right)*(0.2)^0*(1-0.2)^{8-0}\]

\[\Large \left({\color{red}{1}}\right)*(0.2)^0*(1-0.2)^{8-0}\]

Answer 49

after this point, you just need to compute \[\Large \left(1\right)*(0.2)^0*(1-0.2)^{8-0}\] with a calculator to find the value of P(X = 0)

Answer 50

how do you know what p is?

Answer 51

p = 1/5 = 0.2

since the probability of success (ie getting the right answer) is 1/5

1 right answer
5 total choices

Answer 52

oh ok

Answer 53

so do you see how to compute the probability for P(X = 0) ?

Answer 54

I do