Answer:
79.81% probability of winning.
Step-by-step explanation:
For each ticket, there are only two possible outcomes. Either you win, or you do not. The probability of winning in a ticket is independent of other tickets. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
The odds of winning the prize is 1 : 53
This means that ![p = \frac{1}{53}](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B1%7D%7B53%7D)
If I purchased 84 tickets what are the odds of winning in percentage.
To win, at least one of the tickets must have the prize, of 84 tickets, so
.
We have to find
, which is given by:
![P(X \geq 0) = 1 - P(X = 0)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%200%29%20%3D%201%20-%20P%28X%20%3D%200%29)
In which
![P(X = 0) = C_{84,0}.(\frac{1}{53})^{0}.(\frac{52}{53})^{84} = 0.2019](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B84%2C0%7D.%28%5Cfrac%7B1%7D%7B53%7D%29%5E%7B0%7D.%28%5Cfrac%7B52%7D%7B53%7D%29%5E%7B84%7D%20%3D%200.2019)
![P(X \geq 0) = 1 - P(X = 0) = 1 - 0.2019 = 0.7981](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%200%29%20%3D%201%20-%20P%28X%20%3D%200%29%20%3D%201%20-%200.2019%20%3D%200.7981)
79.81% probability of winning.