how can i determine the probability to win one prize lottery where
97 tickets are losing
3 are winning
i have 40 tickets
??
i thought of this:
[math](3 * (97! / (97-94)! ) / ( 100! / (100 - 40)! )[/math]
[math] (3*59*58*57) / (100*99*98) [/math]
yes, im a tard
There are 100 tickets in all and you have 40.
You have a 40% chance of winning AT LEAST one prize.
Does that answer your question or do you need to know the chances of your winning EXACTLY one prize?
I need to know exactly the change of winning only one prize
knowing that there are only three prizes
so far I managed to calculate the possibility when the winning ticket is first or last.. I don't know how to get rid of the '' order ''
get rid of the order by dividing on permutations
what about this, it seems very messy for this simple problem but.
X = getting only 1 winning ticket but at any place
P(X) = 3/100 * 97/99 * 96/98 + ...
which can be written with ! and such , but gets messy, this might be what you did.
then to also take into account where the winning ticket is the 2nd pick or 3th pick. you have to multiply it with the number of all combinations of this
W, L , L ,L .....
L W L L L .... etc.
the number of of these are found with the binomial coefficient
(n, k) = n! / (k!(n-k)!)
bincoeff( 100, 40) * P(X)
Any of 100 tickets could be drawn 1st. You have 40. Chances of winning 1st time out is 40/100, Call that probability A.
Any of 99 could be drawn 2nd. Since we assumed you won 1st time, you can tear up that ticket. It can't be drawn again. You still have 39. Your chance of winning the 2nd draw is 39/99. Your chance of NOT winning is 1 - 39/99. Call that probability B.
Any of 98 could be drawn 3rd. Since you DIDN'T win 2nd draw you still have 39 tickets. Your chance of winning the 3rd draw is 39/98. Your chance of NOT winning is 1 - 39/98. Call that probability C.
A * B * C is the probability you won draw 1 and lost 2 and 3.
Similar analyses of winning ONLY the 2nd draw and ONLY the 3rd draw. Whenever you win, discard winning ticket. If you don't win, you keep same number of tickets. Number of tickets in the hat always decreases after a draw. Multiply chances of winning draw N by chances of NOT winning other 2 draws.
After figuring all 3 cases, add results.
but what if is drawn a ticket of mine that isn't winning?
The hat has numbers 1 to 100 inside.
They draw 3, one at a time, and those are the winners.
They never draw a number and say "this is a loser".
Technically they are saying all the others are losers
3 are drawn. Any number not drawn is a loser.
You can check the answer you get by writing a simulation in some programming language, doing it 100000 times, and then averaging out the results.
The more you do it the closer you'll be to the answer.
This is a method for the retarded ones who don't know how to do math
calculate the probability of not winning on any of the tickets and subtract from 1
your formula doesn't make sense to me
Although this will actually give the probability of winning at least one, and possibly more prizes. Do you want the probability of winning exactly one prize, but not more?
>probability of winning at least one
exactly, I need the probability of only one
the formula I wrote is for calculating the probabilities in case the 1st one is winning and all the others are losing.. but it's not what I need / what the question asks
then what should i do?
OK
im a retard
the chance that a ticket is a winning one is 3%
if I have 40 tickets I only have to sum that changes 40 times
so it's 120% chance
r-right anons?
...
what does this mean?
why doens't anyone reply me?
this is the right answer but you don't multiply P(X) by C(100,40) but rather by C(40, 1) which is just 40
pretty sure anyway
for P(X) you mean
3/100 * D97,40 / D100.40
or
3/100 * 97! / 100!
non of those, P(X) is defined in my post that he replied to
It's hypergeometrical because of multiple draws. So you have sample n=40. You calculate you chances over multiple stages (with each step alter the probability) use the path rule and you end up with your probability.
Use hypergeometric and for winning one prize the probability is 43.7%
this
[eqn]
X \sim \text{Hypergeometric}(40, 3, 100) \\
\begin{align}
\text{P}(X > 0) &= 1 - \text{P}(X = 0) \\
&= 1 - \frac{\binom{3}{0}\binom{97}{40}}{\binom{100}{40}} \\
&= \frac{6374}{8085} \\
&\approx 78.8\%
\end{align}
[/eqn]