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?
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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
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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
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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 $5.
If we are considering the winnings only then
P(5)=4/52
P(A)=4/52
P(Face other than Ace)=(3*4)/52
OpenStudy (dan815):
for the P(A) u win 20 bucks
P(faceothathanAce)=10 bucks
OpenStudy (dan815):
make sense?
OpenStudy (anonymous):
yes
OpenStudy (dan815):
umm okay
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OpenStudy (anonymous):
now what
OpenStudy (dan815):
now to find expected value
OpenStudy (anonymous):
ok
OpenStudy (dan815):
lets see what we got so far
OpenStudy (anonymous):
wait was A correct?
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OpenStudy (dan815):
noh no it was all wrong
OpenStudy (anonymous):
oh okay
OpenStudy (dan815):
i thought u understood...
OpenStudy (anonymous):
give me a sec i was doing two things at once I'm gonna read it over again
OpenStudy (dan815):
okay
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OpenStudy (anonymous):
so it would be
X 5 10 20
P(X) .077 .077 0.23
??
OpenStudy (dan815):
do u understand why?
OpenStudy (dan815):
how many cards in a full deck?
OpenStudy (anonymous):
52
OpenStudy (dan815):
okay how many 2s are in that deck
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OpenStudy (anonymous):
4
OpenStudy (dan815):
ok good so
4/52 = prob of drawing a 2
OpenStudy (dan815):
so 4/52 times u will win 5$ right
OpenStudy (anonymous):
yes
OpenStudy (dan815):
okay now theres a little trick to the next 2
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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
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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?
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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
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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
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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
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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 ??
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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
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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
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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
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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
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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
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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
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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