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Mathematics 11 Online
OpenStudy (anonymous):

need someone to check my work and help with 2/3 part (will medal, fan, and best answer): In a card game (using a standard deck of cards), you pay $5 to draw a single card. If you draw an Ace, you win $20. If you draw a face card, you win $10. If you draw a 2, you win $5. a) Fill in the probability distribution below: X 5 10 20 P(X) .096 2.404 2.788 b) What is the expected value? c) Is the game fair? Why or why not?

OpenStudy (anonymous):

@Zale101

OpenStudy (anonymous):

@dan815

OpenStudy (dan815):

did u do a)?

OpenStudy (anonymous):

yes its the values plugged in for P(x) they were originally blank, could you check if i did it correctly?

OpenStudy (dan815):

ok i have to use the washroom ill brb

OpenStudy (anonymous):

ok

OpenStudy (dan815):

sry got distracted

OpenStudy (dan815):

okay lets do this problem of yours

OpenStudy (dan815):

okay now i have a question for you

OpenStudy (dan815):

remember that you are paying 5 dollars to buy in so all your winnings u have to subract 5 dollars dont u

OpenStudy (dan815):

In a card game (using a standard deck of cards), you pay $5 to draw a single card. If you draw an Ace, you win $20. If you draw a face card, you win $10. If you draw a 2, you win $5. If we are considering the winnings only then P(5)=4/52 P(A)=4/52 P(Face other than Ace)=(3*4)/52

OpenStudy (dan815):

for the P(A) u win 20 bucks P(faceothathanAce)=10 bucks

OpenStudy (dan815):

make sense?

OpenStudy (anonymous):

yes

OpenStudy (dan815):

umm okay

OpenStudy (anonymous):

now what

OpenStudy (dan815):

now to find expected value

OpenStudy (anonymous):

ok

OpenStudy (dan815):

lets see what we got so far

OpenStudy (anonymous):

wait was A correct?

OpenStudy (dan815):

noh no it was all wrong

OpenStudy (anonymous):

oh okay

OpenStudy (dan815):

i thought u understood...

OpenStudy (anonymous):

give me a sec i was doing two things at once I'm gonna read it over again

OpenStudy (dan815):

okay

OpenStudy (anonymous):

so it would be X 5 10 20 P(X) .077 .077 0.23 ??

OpenStudy (dan815):

do u understand why?

OpenStudy (dan815):

how many cards in a full deck?

OpenStudy (anonymous):

52

OpenStudy (dan815):

okay how many 2s are in that deck

OpenStudy (anonymous):

4

OpenStudy (dan815):

ok good so 4/52 = prob of drawing a 2

OpenStudy (dan815):

so 4/52 times u will win 5$ right

OpenStudy (anonymous):

yes

OpenStudy (dan815):

okay now theres a little trick to the next 2

OpenStudy (dan815):

How many face cards are there?

OpenStudy (anonymous):

12 and two jokers

OpenStudy (dan815):

2 jokers are not included in a standard deck

OpenStudy (anonymous):

okay so just 12

OpenStudy (dan815):

there are actually 16 face A,,J,Q,K but they say for the 4 Aces its 20 and 4 Jacks,4 Queens, 4 Kings its 10

OpenStudy (dan815):

okay so u get this stuff now, lets move on, do u know how to take average?

OpenStudy (anonymous):

how to get the average?

OpenStudy (dan815):

okay like if i say 1 ,2 ,3 u have these 3 numbers what is the average of these 3 numbers?

OpenStudy (anonymous):

oh yeah i know how to do that, 2 is the avg

OpenStudy (dan815):

okay heres a harder one how about 2,5,3,1 whats avg?

OpenStudy (dan815):

u can give it to me in a fraction

OpenStudy (anonymous):

2 3/4

OpenStudy (dan815):

nice! or u can just say 11/4

OpenStudy (dan815):

u did (2+5+3+1)/4 right?

OpenStudy (anonymous):

yes

OpenStudy (dan815):

i want to show u a slightly different way of doing it and that is how we will find the answer to ur problem

OpenStudy (anonymous):

okay

OpenStudy (dan815):

now if i were to write 1,2,3 like in sticks

OpenStudy (dan815):

1,2,3 l, ll ,lll

OpenStudy (anonymous):

ok

OpenStudy (dan815):

if someone were to tell you to rearrange these sticks so all 3 have the same

OpenStudy (anonymous):

two in each

OpenStudy (dan815):

l,ll,lll ^--- you can take one stick from here and stick in the first one ll,ll,ll and you get this

OpenStudy (anonymous):

yes

OpenStudy (dan815):

okay this is basically what avferage is all about

OpenStudy (anonymous):

okay

OpenStudy (dan815):

if someone gave u by numbers with all differen quantities they just want to know what same 5 numbers can u put 5 times so they all add up to the same

OpenStudy (dan815):

and sometimes u have to break the sticks in halfs or 1/3rds and so on, so that it works out

OpenStudy (dan815):

now this is basically the same question they are giving u

OpenStudy (dan815):

so it would be X 5 10 20 P(X) .077 .077 0.23 ??

OpenStudy (dan815):

lets take a look at this

OpenStudy (anonymous):

wait just to be clear, we're working on b now right, was a correct? (just making sure)

OpenStudy (dan815):

yeah

OpenStudy (anonymous):

okay

OpenStudy (dan815):

X 5 10 20 P(X) .077 .077 0.23 ?? lets break this problem down in the same thing

OpenStudy (anonymous):

you want me to find the avg of the three numbers?

OpenStudy (dan815):

slightly different here

OpenStudy (dan815):

first lets complete our set

OpenStudy (anonymous):

okay

OpenStudy (dan815):

we saw how hes winning right but how many cards are ther where he gets 0 dollars

OpenStudy (anonymous):

still 52

OpenStudy (dan815):

i mean out of the 52 cards, how many cards give him no money

OpenStudy (dan815):

let see how many winnings cards are there lets count them

OpenStudy (anonymous):

oh 28

OpenStudy (dan815):

we have four 2s four As twelve other faces

OpenStudy (anonymous):

20 i mean

OpenStudy (dan815):

okay so 20/52 give u money so how many dont give u money then?

OpenStudy (anonymous):

32

OpenStudy (dan815):

okay good so lets write that in too

OpenStudy (dan815):

X 0 5 10 20 P(x) 32/52 4/52 12/52 4/52

OpenStudy (anonymous):

ok

OpenStudy (dan815):

make sense so far, looks like u put in a wrong number for part A correct it

OpenStudy (dan815):

5 and 20 shud have same probabililty under them and 10 shud have 3 times the prob

OpenStudy (anonymous):

oh right my bad sorry but yes makes sense a should be X 5 10 20 P(X) .077 .0.23 .077

OpenStudy (dan815):

okay good now let me show u a small trick we will use

OpenStudy (dan815):

u saw how 1,2,3 avg was 2

OpenStudy (dan815):

if u have 1,2,3, 1,2,3 <--- 2 of these the average of these 6 numbers i claim is still 2

OpenStudy (dan815):

can u see why taht wud be true

OpenStudy (anonymous):

yes

OpenStudy (dan815):

okay in the same way

OpenStudy (dan815):

i can even have 52 sets of them and it will still be 2

OpenStudy (anonymous):

yes

OpenStudy (dan815):

1,2,3 , 1,2,3 ,1,2,3 ... because ill always be able to do that trick of moving the stick into one of the 1s and make it all be 2

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