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/sci/ - Science & Math

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2838711 No.2838711 [Reply] [Original]

182% of Americans cant get this right.

>> No.2838724

Is it 25%?

>> No.2838745

33%

>> No.2838748
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2838748

>> No.2838753

50%
I dont give a shit if I'm wrong

>> No.2838756

50%

>> No.2838762

50%

>> No.2838766

So you're saying only 18% of people in the entire world get this right? 100% of americans get it wrong and 82% of non americans get it wrong


50/50

>> No.2838771

I remember the days of the good troll threads.

Like that multiple choice question about the probability of you selecting the right answer.

Wait, I don't miss that at all.

>> No.2838765

flip a coin 3 times: 4 possible outcomes
2 heads and a tail: 1 outcome
1/4 or 25% first guy was right

>> No.2838778

Lol @ the dude that fail logic'd and deleted his post. I think you worry about what people think about you too much.

>> No.2838785

37.5%

>> No.2838792

Khan academy -> probabilities 1

>> No.2838796

what would be said probability?


and what would be a good statistic text to stop being so dumb?

>> No.2838797

P = sum([n!/(k!*(n-k)!)]*(1/2)^k*(1 - 1/2)^(n-k)) with n = 3, 2<=k<=3

P = 1/2

>> No.2838800

>>2838792
Fuck off you cunt I went to Jedi Academy. Eat shit.

>> No.2838806

its 50%

>> No.2838818

>2838797
It took an americunt engineer to explain it properly to you guys

>> No.2838831

>>2838785
Bingo. The question is asking the probability of it landing on heads at least two times. So you take the probability of it landing on heads twice (.25) and the probability of it landing thrice (.125), and you add those two to get the answer.

Everybody else is retarded and a troll.

>> No.2838832

12.25
>0.5*0.5*0.5
fuck yeah year 10 maths!

>> No.2838854

>>2838831
you are not completely correct. There are 3 ways for it to land on heads twice. Refer to >2838797
That is the correct answer.

>> No.2838850

>>2838832
just reread the question. .
0.25+12.5
P(2heads)+P(3heads)

>> No.2838870

>>2838831
No. Use a binomial distribution, like >>2838797
It is you who is retarded and a troll.

>> No.2838876

Let me break it down for you guys. There are 4 possible events that satisfy OPs question.
1) heads, tails, heads P_1 = (1/2)*(1 - 1/2)*(1/2) = 1/8)
2) heads, heads, tails P_2 = (1/2)*(1/2)*(1 - 1/2) = 1/8
3) tails, heads, heads P_3 = (1 - 1/2)*(1/2)*(1/2) = 1/8
4) heads, heads, heads P_4 = (1/2)*(1/2)*(1/2) = 1/8
P = P_1 + P_2 + P_3 + P_4 = 4*(1/8) = 1/2
>schooled by engineer suckas

>> No.2838889

binomial(2,3,0.5) + binomial(3,3,0.5)
= 0.5

>> No.2838900

Its not really 50% right?


Can someone explain that to me.

>> No.2838901

>>2838876
This. P=.5

>> No.2838913

>>2838900
it's already been explained. P = 1/2 = 0.5 which is the same as saying P = 50%

>> No.2838922

50%

The outcome of each flip does not affect the outcome fo the next flip. Each flip is a separate event and the probability stays at 50% for each.

old troll is old, yet people still fall for it

>> No.2838934

You don't even need fancy maths. Basic arithmetic and reason will do.

Each coin can land on either heads or tails.
If the probability of of it landing on heads twice was less than 50%, that means the probability of it landing tails twice would be MORE than 50%.
And since you know heads is just as likely as tails, this cannot be right.

>> No.2838938

>>2838900
HHH
HHT
HTT
TTT
4 outcomes. 2 with 2+ heads.

>> No.2838957

>>2838938
hurr durr
HTH
Duererererrdd

>> No.2838960

P={1:3, 1:2}

>> No.2838966

Guys can you help me with this?

You roll a dice 3 times, what is the chance of getting one or more 5's?

>> No.2838973

>>2838797
not sure if trolling or just retarded

>> No.2838985

>>2838922
your double trolling
because theres 50% for the first one
and then another 50% for the second one

so 25%?

>> No.2839016

>>2838960
ok, lemme brake it down for you.
3*{H:T}
{H}{H}{H:T}
{H, H, H} = 3/6 = 1:2
{H, H, T} = 2/6 = 1:3
P = {1:3, 1:2}

>> No.2839081

>>2838985
Really? You must be really retarded.
The law of averages does not apply in coin flips. You have to remember that each coin flip is independent.

The answer will be 50% no matter how many flips. Look up independence and coin flips.

Fucking retarded high schoolers

>> No.2839120

>>2839081
you're wrong
see
>>2839016

>> No.2839131

>>2839120
Again, that doesnt take into account that each coin flip is independent.


Look up independence... Then go take a statistics class.. then you can go kill yourself from the shame.

>> No.2839145

>>2839131
You should really delve a little deeper into statistics, you're right in saying that each coin flip is independent of the last, but when asked as if the coins were flipped in succession and the point of being asked is before the initial flip, 1 coin flipped 2 times is still 1/4 to guess each landing correctly ( including the number of which flip it landed on each face. ), not 1/2.

>> No.2839152
File: 15 KB, 113x290, hardesteasy.jpg [View same] [iqdb] [saucenao] [google]
2839152

>>2838771
Fuck that, this was classic though. Find x using only the basic axioms of Euclidean geometry, i.e. no law of sines, pythagorean theorem, etc. inb4 "indeterminate"

>> No.2839169

50%

>> No.2839171

HHH
HHT
HTH
HTT

TTT
TTH
THT
THH

4/8 = 1/2

>> No.2839175

Here are the possible outcomes of three flips:

TTT
TTH
THT
THH
HTT
HTH
HHT
HHH

There are eight total outcomes. Four of these involve two or more heads flips. Unless I'm derping hardcore, the answer is 1/2.

>> No.2839189
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2839189

>>2839152
y'all here ya go this problem was up a few weeks ago

>> No.2839200

>>2839131
http://arnoldkling.com/apstats/coins.html
independence does not dictate the probability of outcomes

>> No.2839207
File: 234 KB, 1209x706, You should be able to solve this 42236054.jpg [View same] [iqdb] [saucenao] [google]
2839207

数学しなイカ？

>> No.2839209

>>2839175
>>2839171
derping
see
>>2839016

>> No.2839241
File: 34 KB, 598x350, problembility.jpg [View same] [iqdb] [saucenao] [google]
2839241

>>2839207

>> No.2839245

>>2839207
this is the same thing as flipping one coin right?
lemme break down into logic/set-theory:
2{H:T}, {H} |- {H, H}
{H}/\{H:T}
{H:T}; P={1:2}

>> No.2839255

>>2839241

>nothing_to_geso_about.jpg

>> No.2839328

>>2839241
|- {H, H}
10{H:T}
3{H}7{H:T}
-5{H:T}=2{H}3{H:T}
-3{H:T}=2{H:T}
P={1:2}

Now that's what I call a good display of independence.

proof read me someone?

>> No.2839519

>>2839241
224.0/791.0
0.2831858407079646

Brute forced.

>> No.2839523

{HHH, HHT, HTH, HTT, THH, THT, TTT}
P(HHH \/ {HHT, HTH, THH})
=P(HHH)+P(HHT, HTH, THH)
P(HHH)=1/6
P({HHT, HTH, THH})=3/6
P(HHH \/ {HHT, HTH, THH})=2:3

>> No.2839538

Four ways to get is right,
H H H
H H T
H T H
T H H

four ways to get is wrong:

H T T
T H T
H T T
T T T

Then there is the probability that the thrower quits before being bothered to finish, or loses the coin down a drain, or forgets what he got on the throws.

Probability of two or more heads VERY SLIM

>> No.2839599

>>2839538
What did I do wrong then?
>>2839523

>> No.2839614

>>2839538
>mfw you put HTT twice

>> No.2839646

>>2839523
>mfw I forgot to put TTH and fucked up my ratios
lemme try this one last time:
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
P(HHH, HHT, HTH, THH)
=1:2

Ok, fine. You guys were right. I'm just retarded.

>> No.2839686

>>2838711
After staring at my fingers for a minute or so I'm gonna say 50%.

>> No.2839711

(3!/(2!(1!))).5^3

Why do threads like this get 50 posts? are you physics majors so incompetent that you freeze at the sight of finite math?

>> No.2839728

>>2839711
sorry, I'm a chem major. shit doesn't make sense.
can you break this down for me in layman's?

>> No.2839734

>>2839728
http://en.wikipedia.org/wiki/Binomial_distribution

>> No.2839754

>>2839711
This gets you exactly two heads, not at least two heads.

>> No.2839766

Think of it like this:

For every possible distribution of heads and tails, there is a distribution which is exactly opposite of the one you picked.

So if you have H, the opposite is T. Therefore, picking H is a 1/2 chance, because each choice you make has an opposite "pair" with the same heads/tails value.

If you flip it twice, you have HT, TH, HH, and TT for possible outcomes. Since half of these outcomes have the same heads/tails value (ie, one of each, or two of each), you're really only picking between 2 outcomes, thus making the probability 50% no matter how many times you flip the coin, or how many specific heads or tails you ask for.

>> No.2839782

>>2838711

8 possibilities:

0 heads
first, second or third is heads
first two, last two, first and last are heads
all of them are heads

3 of those are winning situations. Therefore, chance is

3/8

>> No.2839786

>>2839782

wait, didn´t read the "or more". Then there are 4 lucky cases what makes the chance 4/8 = 1/2

>> No.2839800

>>2839754
(3!/(2!(1!))).5^3 + .5^3

>> No.2840320
File: 320 KB, 601x706, 1301449546619.png [View same] [iqdb] [saucenao] [google]
2840320

You flip 4 coins. What's the probability that at least one of them is tails? Describe the solution as a sum of probabilities.
(not a homework)

>> No.2840335

>>2840320
1-1/16=15/16.
And fuck you, I will describe solution how I want.

>> No.2840368

50%

>> No.2840373

>>2839538
he didn't say anything about particular order you fuckwit

>> No.2840400

>>2840373
It really doesn't matter.
If you don't take order into account you have 4 possible outcomes: 0,1,2 or 3 heads. The probability of 2 or 3 heads is 2/4=50%.

>> No.2840427

>>2840400
True, but now you need to prove that (or at least explain why) all the four outcomes have the same probability.

>> No.2840440

>>2840335
ok then how about 1245 coins and 354 heads?

>> No.2840466

hhh
hht
hth
htt*
thh
tht*
tth*
ttt*

>> No.2840476 [DELETED] 

>>2840440
354/(1245^2)

>> No.2840481

>>2840440

I think, something like <span class="math">\Phi(345/\sqrt{1245})[/spoiler], or about 1-10^(-22)

>> No.2840489

>>2840481
No, wait, it's actually <span class="math">\Phi(277.5/\sqrt{1245})[/spoiler], or just 1-10^(-15)

>> No.2840697

>>2839241
~17.992%

the solution:
first we have 10 flips 3 of witch we know are heads.
looking at the rest of the task we see that we need at least one more heads to find the answer.
the answer of this question i found on the net it's too much computation for me to handle but if anyone's interested:
http://mathforum.org/library/drmath/view/52217.html
now we know that the probability that at least one more coin is heads is 53/64

now our friend takes 5 coins 2 of witch we know are heads
so the probability that he takes another coin that is heads is (2/8+2/7+2/6)=0.869048

after that we take 3 coins from him and one of them we know is heads.
so the probability that the other 2 we took are tails is 1/4

and finally the probability that his coins are heads is:
(53/64)*0.869048*(1/4)=~0.17992=~17.992%

or maybe i'm wrong and wasted an hour of my life

>> No.2840703
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2840703

>>2840697
>an hour
I'd be ok with this if you were just playing around with math but an hour?

Shame.

>> No.2840741
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2840741

>>2840697
fuck.. it's wrong
disregard this post