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/sci/ - Science & Math

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File: 51 KB, 457x456, Loonie.jpg [View same] [iqdb] [saucenao] [google]
2151919 No.2151919 [Reply] [Original]

Hey /sci/entists.

You flip 6 coins.

What is the probability that 3 coins land as heads and 3 coins land as tails?

The coins can be in any order.

>> No.2151928

1/2.

/thread

>> No.2151938

3/6, so 1/2?

>> No.2151944

1/8.

/thread

>> No.2151945

6/6 The coins have 2 heads

>> No.2151946

That doesn't make any sense.

>> No.2151956

1/2^3 so 1/8 try harder

>> No.2151958

Let X be the amount of heads

X~Bin(6, 0.5)
P(X=3)=6C3*0.5^3=0.3125=5/16

>> No.2151962

>>2151958

I do not understand this.

>> No.2151984

>>2151962
>he doesn't understand discrete random variables
>laughing statisticians.jpg

>> No.2152007

>>2151984
I don't know why I laughed at this.

>> No.2152022

>>2152007
You laughed because it was funny obviously

>> No.2152033

I just flipped 6 pennies

5 landed heads up
1 landed tails up

where is your god now?

>> No.2152059

>>2152033
P(X=5)=6C5*0.5^6=0.09376=3/32

still not exactly unlikely to happen

>> No.2152060

>>2152059
whoops,
>0.09375

>> No.2152063
File: 35 KB, 812x612, coin tosses.png [View same] [iqdb] [saucenao] [google]
2152063

Everyone fucking failed probability

You must include degeneracies when talking about probabilities.

>> No.2152067 [DELETED] 

>>2151962

Basically the formula <span class="math">P(X=k) = \binom nk p^k (1-p)^{n-k}[/spoiler]
is the probability of something happening <span class="math">k[/spoiler] times in <span class="math">n[/spoiler] repetitions where the odds of a single event is<span class="math">p[/spoiler]

So in this case <span class="math">\binom{6}{3} 0.5^3 (1-0.5)^{6-3}[/spoiler]

>> No.2152071

Chance to get 3 coins head and 3 tails is (1/2)^6 = 64, there are 20 different ways to get it so 20/64 = 5/16.

>> No.2152091

>>2151962

Basically the formula <span class="math">P(X=k) = {n\choose k} p^k (1-p)^{n-k}[/spoiler]
is the probability of something happening <span class="math">k[/spoiler] times in <span class="math">n[/spoiler] repetitions where the odds of a single event is<span class="math">p[/spoiler]

So in this case <span class="math">{6\choose3} 0.5^3 (1-0.5)^{6-3}[/spoiler]

>>2152063
Get the fuck out of here. Occam's razor, ever heard of it?

>> No.2152092

>>2151958

You used the binomial distribution incorrectly.

You have to multiply by the probability that heads does not happen as well as that it does happen.

Correct binomial probability: 6C3 x 0.5^3 x 0.5^3 = 6C3 x 0.5^6 = 5/16

Well, you got the same answer as me but you just wrote out your working wrong.

>> No.2152097

It's 5/16

Number of total outcomes is 2^6 = 64

Number of ways to achieve 3 heads 3 tails is 6C3 (6 choose 3) = 20

20/64 = 5/16

>> No.2152098

>>2152091
>Get the fuck out of here. Occam's razor, ever heard of it?

Of course I have, the pic was from a previous iteration of this troll thread where someone was claiming you were just as likely to get 99heads and 1 tail as you were to get 50head and 50 tails.

>> No.2152103

>>2152092
I simplified my working because p=(1-p)

>> No.2152107

>>2152103
in this specific case anyway.

>> No.2152111

>>2152103

Yeah but you wrote 6C3 * 0.5^3 rather than 6C3 * 0,5^6. Doesn't matter, was just a small slip.

>> No.2152126

>>2152111
whoops, thanks for pointing that out to me, I did mean ^6, because thats what I worked out.

>> No.2152200

5/16

/THREAD

>> No.2152222

Not 50 50, simply because one side may have a heavier mass because of the emblem or description, which would weight it down, resulting in oddities.