Hey...

I need help with homework

1) If you deal the top three cards from a 52-card deck, what is the probability the cards are two clubs followed by a heart?

2) The probability of rolling five 6s from a cup with five dice is .0001286. If you shake the cup 3 times a day, every day, for 20 days in a row, what is the probability of rolling five 6s at least once?

3) If player A's free throw percentage is 63%, and player B's is 82%, what's the probability of player A making two consecutive free throws followed by B making two consecutive free throws?

And, yes I did try them already. For #1, I got .01529, and for #3, I got 0.2668. Can anyone offer any help please? Thank you!
i honestly can't see people helping other people with math online....
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i honestly can't see people helping other people with math online....

Especially in the pit.
darkstar would if he was online, he's our resident maths genius.
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Hey...

I need help with homework

1) If you deal the top three cards from a 52-card deck, what is the probability the cards are two clubs followed by a heart?

39/52?
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If this even madkes sense... if yhou sig this, Iw ll kill you.
number 1's a trick question. A new 52 card deck is organized Aces, 2s, 3s :P
When all else fails, ask the pit.
the first question is 1/132600, although i sucked at probabilty i thnk you either times or add, i times:
1/52 x1/51 x1/50=1/132600
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****ing math. you don't need math in real life
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1) If you deal the top three cards from a 52-card deck, what is the probability the cards are two clubs followed by a heart?

Very low.

3) If player A's free throw percentage is 63%, and player B's is 82%, what's the probability of player A making two consecutive free throws followed by B making two consecutive free throws?

Depends on the lucky bounces.
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3) If player A's free throw percentage is 63%, and player B's is 82%, what's the probability of player A making two consecutive free throws followed by B making two consecutive free throws?

Depends on the lucky bounces.

And the ability of your dice hand. CLAK CLAK. gotta have a good dice hand.
Quote by ZanasCross
I'm now so drunk that even if my mom had given me a blow job at aeg 2, i'd be like I'm a pmp, butches.!

If this even madkes sense... if yhou sig this, Iw ll kill you.
I'm pretty sure the first is 13/52*12/51*13/50. How many times you shake the cup is irrelevant btw. I think its .0001286*20 but I have a sneaking suspicion that's wrong though. The third is .63*.63*.82.*.82
EDIT: Nvm, thanks. First and third make sense.

Last edited by kirbyrocknroll at May 21, 2008,
Quote by rjdusa
I'm pretty sure the first is 13/52*12/51*13/50

This is correct for the first.

Explanation:

13 clubs, 52 cards. Draw one. 12 clubs left, 51 cards left. Draw One. 13 heart cards, 50 cards left.
IIIIIII
Last edited by synopsize at May 21, 2008,
Quote by kirbyrocknroll
How'd you get it?

39/52? How?

I dunno, Each suit has 13 cards, so the chance of drawing a heart is like, 13/52? the chance of 2 clubs is 26/52? im not to good at maths
Quote by ZanasCross
I'm now so drunk that even if my mom had given me a blow job at aeg 2, i'd be like I'm a pmp, butches.!

If this even madkes sense... if yhou sig this, Iw ll kill you.
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****ing math. you don't need math in real life

What about when you go to the grocery store and the cashier charges you more than he's supposed to? You should at least know how to add and multiply. Math is useful
Quote by rjdusa
I'm pretty sure the first is 13/52*12/51*13/50. How many times you shake the cup is irrelevant btw. I think its .0001286*20 but I have a sneaking suspicion that's wrong though. The third is .63*.63*.82.*.82

im a math major and i took stats 2 simesters ago and i say thats pretty dead on right there
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Quote by kirbyrocknroll
How'd you get it?

39/52? How?

no ignore me im really wrong its going to be

13/52 x 12/52 x 13/52 =0.014

i think, god im doing A-level math an dim retarded
Quote by Mathamology
One day that guy is gonna lose a whole arm to that blender

and that shall be the day I laugh the hardest
Quote by kirbyrocknroll
Hey...

I need help with homework

1) If you deal the top three cards from a 52-card deck, what is the probability the cards are two clubs followed by a heart?

P(club) = 1/4.
P(club | club) = 12/51
p(heart | not heart so far) = 13/50

multiple them together. 0.01529... like you said

2) The probability of rolling five 6s from a cup with five dice is .0001286. If you shake the cup 3 times a day, every day, for 20 days in a row, what is the probability of rolling five 6s at least once?

so 60 throws? bionomial distribution i guess.

n= 60
p = (1/6)^5.

and P(x >=1) = 1 - P (x=0)

basically, the chance of it happening once or more in 60 throws is 1 - the probability it won't happen at all. so using the binomial working out thing you get 1 - 0.9923... which is 0.007686...

3) If player A's free throw percentage is 63%, and player B's is 82%, what's the probability of player A making two consecutive free throws followed by B making two consecutive free throws?

as all the throws are independant you can go ahead and multiply it all together. which comes out at 0.2668...like you got.

k.
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...that I am cursed far more than I am praised."
Last edited by Sol9989 at May 21, 2008,
Quote by Opethfan1
im a math major and i took stats 2 simesters ago and i say thats pretty dead on right there

I'm not taking stats until next year either .
i would think that b) the probability would stay the same.

As the odds wouldnt change.
Quote by rjdusa
I'm pretty sure the first is 13/52*12/51*13/50. How many times you shake the cup is irrelevant btw. I think its .0001286*20 but I have a sneaking suspicion that's wrong though. The third is .63*.63*.82.*.82

Thanks I don't get what you mean for #2 though. How is the number of shakes irrelevant? Aren't you more likely to roll 5 6s if you try 60 times, rather than 1? Unless of course you meant each actual shake before rolling By "shake" I meant roll. So you roll it 3 times a day for 20 days.
Quote by kirbyrocknroll
Thanks I don't get what you mean for #2 though. How is the number of shakes irrelevant? Aren't you more likely to roll 5 6s if you try 60 times, rather than 1? Unless of course you meant each actual shake before rolling By "shake" I meant roll. So you roll it 3 times a day for 20 days.

see my answer. if you've done about the binomial distribution and/or bernoulli trials it should make sense. the number of throws is relevant, but with probabilities that small, it's kind of diminished relevance.
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lol im going to be a senior in pre-algebra. im pretty much a complete math retard... and im ok with that.
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Quote by kirbyrocknroll
Thanks I don't get what you mean for #2 though. How is the number of shakes irrelevant? Aren't you more likely to roll 5 6s if you try 60 times, rather than 1? Unless of course you meant each actual shake before rolling By "shake" I meant roll. So you roll it 3 times a day for 20 days.

I took it as shakes before rolling, like if you shake the dice in the cup 3 times then roll them. I think one of the people above was right when saying it would be the odds of doing once since the odds wouldn't change. It would go from 1/7776 to 2/15552 which is still 1/7776. I wasn't sure if it was that or the other way I did it at first but the more I look at it the more I think this way is right.
Oh, okay. Thanks everyone!

EDIT: Sol, how did you get .9923 for P(none) for #2?
Quote by kirbyrocknroll
I don't understand why you multiply 1/7776 by 60 for #2

you don't. simple.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
I still don't quite get it. Could someone please go step by step for #2?