Chances / Probability

[Charles Pegge]

Code Select Expand

macro Chances as float(r, n,p)
==============================
'n nomber of opportunities or trials
'p probability of a single coincidence
float np=1.0-p
r=pow(np,n)
r=1.0-r
end macro
'
'TEST
'going to a party with 50 guests.
'what is the chances that someone at the
'party has the same birthday as you?
print str(chances(50, 1/365)*100,0) "%"  '13%
'also
print str(chances(300, 1/365)*100,0) "%" '56%


[Nicola]

I asked Copilot, and it told me that if I and 49 other people are invited to the same party, the probability that there is another person with the same birthday as mine is 12.5%.

[Charles Pegge]

It is 1 - (the probability of not having the same birthday as any single person) ^ 49


1.0 - pow( (364/365), 49)

[Nicola]

💪

[Charles Pegge]

Here is another:

8 snooker balls are contained in a bag. Only 3 balls are colored red. What is the probability that when you randomly pick 3 balls out of the bag, they are all red?

[Nicola]

Charles, 
I can't apply your macro to solve this problem

[Theo Gottwald]

Then just don't go to that party.
Send Jose instead.

[Charles Pegge]

8 snooker balls are contained in a bag. Only 3 balls are colored red. What is the probability that when you randomly pick 3 balls out of the bag, they are all red?

No, the macro does not fit the task.

One solution:

(3/8) * (2/7) * (1/6) =

1/56

[Zlatko Vid]

 

[Charles Pegge]

factorials can also be used to solve this problem:

( 3! * 5! ) / 8!

which expands to:

( 2 * 3 * 2 * 3 * 4 * 5 ) / ( 2 * 3 * 4 * 5 * 6 * 7 * 8 )

which reduces to:

1/( 7*8 )

1 / 56

[Nicola]

Hi,
combinations of n items chosen k at a time

Code Select Expand

'Le combinazioni rispondono a questa domanda:
'In quanti modi posso scegliere k elementi da un insieme di n elementi, senza ordine e senza ripetizione?
'
'Esempio classico:
'"Quanti gruppi da 3 persone posso formare con 8 persone?"
'Qui l'ordine non conta:
'il gruppo (A,B,C) è lo stesso di (C,B,A).
'---------------------------------------------------------------

' Calcola le combinazioni C(n, k) come float
function Comb(n as int, k as int) as double
    int i, kk
    double num = 1.0
    double den = 1.0

    ' Symmetry: C(n,k) = C(n,n-k)
    kk = k
    if kk > n - kk then kk = n - kk

    if kk < 0 or kk > n then
        return 0
    elseif kk = 0 or kk = n then
        return 1
    end if

    for i = 1 to kk
        num *= (n - kk + i)
        den *= i
    next

    return num / den
end function


'nel caso ci sono 3 palline rosse in un gruppo di 8 palline
'la probabilità che si riesca a pescare 3 palline tutte rosse
'viene calcolata
float num, den
num=Comb(3, 3) 'c'è un solo modo per prendere tutte e 3 rosse
den=Comb(8, 3) ' 56 è il numero delle cominazione totali a 3 a 3
double p = num/den * 100

print "P(all red) = " p