[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


View post   

File: 3 KB, 321x302, prob_coin.gif [View same] [iqdb] [saucenao] [google]
4751568 No.4751568 [Reply] [Original]

i don't understand how to calculate probabilities with tree diagrams. how do i do that?
>the picture is a coin that one flips, with two possible outcomes within each flip. Easy stuff right there
how does one find the probability of rolling two heads in a row? or maybe a heads, heads, and then tails?

>> No.4751595

you assume that all possibilities are equally probable, so to find the probability of something, you just pick the number of favorable cases and divide between the total number

>> No.4751596

Count the number of outcomes (8 in your pic)
Count the number of outcomes that matches the event.
The probability of the event happening is the ratio of these two numbers.

>> No.4751597

I'm going to assume you're overthinking it. You have a list of possible outcomes. Some of those are the desired outcome. (number desired outcome)/(total number of outcomes) = (probability of desired outcome)

>> No.4751600

http://www.mathplayground.com/

>> No.4751624

>>4751600
harharhar
>>4751595
ok so what is the experimental probability of rolling a H, T, and H?

>> No.4751633

bump

>> No.4751651
File: 18 KB, 160x160, frynotsureiftroll.jpg [View same] [iqdb] [saucenao] [google]
4751651

>this thread

You just count them....

>> No.4751660

>>4751651
>experimental probability
did you read the op?

>> No.4751675

>>4751660
>did you read the op?

Yeah....

>> No.4751689
File: 993 KB, 236x224, 1337305254642.gif [View same] [iqdb] [saucenao] [google]
4751689

>>4751675
ok so tell me how you just simply count the experimental probability, please

>> No.4751705

>>4751689

The first three replies all already answered you.

>> No.4751710

>>4751624
In this order, the probability is 1/8.
If you want two heads and one tail in any order, the probability is 3/8.

You can say you have 1/2 probability of getting H and 1/2 probability of getting T. So you have 1/2*1/2 = 1/4 probability of getting HT, 1/4 of getting TH, 1/4 of getting HH, 1/4 of getting TT.

The probability of getting HTTHHTHTHTTTTH in this order is (1/2)^14 because you have 14 coin flips.

>> No.4751726

>>4751710
ok i understand now. thank you very much, as i was struggling to understand it.
>>4751597
>I'm going to assume you're overthinking it.
you were definitely right when you said that, now that i realize all that has to be done!

>> No.4751744 [DELETED] 

>>4751710
i think you meant
>the probability is 1/8.

>> No.4751776
File: 28 KB, 738x357, amazing1.png [View same] [iqdb] [saucenao] [google]
4751776

This thread actually got me thinking, what would be the equation to find any sort of pattern for any number of coin flips? E.g. if you 4 flips and wanted any combination of H, H, H, T? Or H, H, T, T? Or if you restricted the pattern to having H first and last and any combination in the middle? Or any combination with at least 1 T?

>> No.4751782

>>4751776
Formula*