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Mathematics 13 Online
OpenStudy (anonymous):

For fun Suppose you roll a die, if you get 1 and 2, i give you 200$, but if you get 3,4,5,6 you give me 100$. whats the chance that you would lose any money if u rolled the die 10 times

OpenStudy (anonymous):

I never lose.

OpenStudy (anonymous):

What are the chances of me rolling a die to lose? How many possibilities out of 6?

OpenStudy (anonymous):

im asking :P

OpenStudy (anonymous):

No you asked a question. I am asking you a part of your question. You should know at least this.

OpenStudy (anonymous):

i know how to solve it :O just for fun

OpenStudy (anonymous):

-.-

OpenStudy (anonymous):

xD

OpenStudy (anonymous):

The answer should be 20/3 but i might be wrong.

OpenStudy (anonymous):

you are wrong for one reason 20/3 >1

OpenStudy (anonymous):

I told you I was wrong The chances of me rolling a die to lose is 4/6 == 2/3 I don't know how to account for it 10 times

OpenStudy (anonymous):

hmm well think again 2/3 is high value, also its binomial distribution .... try it :)

OpenStudy (anonymous):

I honestly don't know. Show me yourself.

OpenStudy (anonymous):

i wont reveal lets see if others can do it ^_^

OpenStudy (anonymous):

remember my question is asking about losing money, (not guessing wrong)

OpenStudy (anonymous):

haha, well the cute thing is the chance is pretty law so if it was gambling i'll play this game :D

OpenStudy (rational):

expected value = 0

OpenStudy (anonymous):

hmmm interesting ...

OpenStudy (anonymous):

but no xD

OpenStudy (rational):

out of 10 rolls, you need to loose at least 7 times to make negative money

OpenStudy (anonymous):

yessss exactly xD

OpenStudy (rational):

simply find P(x >= 7)

OpenStudy (anonymous):

yesss !

OpenStudy (anonymous):

OHHHH

OpenStudy (anonymous):

You can do a probability tree to find out the possibilities of losing

OpenStudy (anonymous):

:D

OpenStudy (rational):

that doesn't look right, one sec

OpenStudy (anonymous):

hehe see its small value

OpenStudy (rational):

binomial distribution : \[P(X\ge 7) = \sum\limits_{k=7}^{10}\binom{10}{k} (4/6)^{k} (2/6)^{10-k} = \dfrac{11008}{19683} \approx 0.559 \]

OpenStudy (anonymous):

it should be 1- ur value right ?

OpenStudy (rational):

\(0.559 = 55.9\%\) chance for loosing money

OpenStudy (rational):

thats not small, i wouldn't be willing to play this game wid u lol

OpenStudy (anonymous):

hehe ok i made same mistake u did id first reply :P

OpenStudy (anonymous):

would u play tossing a coin ?

OpenStudy (rational):

it is wise not to play any casino games wid statisticians :P

OpenStudy (anonymous):

lol why so

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