Respuesta :
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The odds of winning the prize is 1 : 53
This means that [tex]p = \frac{1}{53}[/tex]
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 [tex]n = 84[/tex].
We have to find [tex]P(X \geq 0)[/tex], which is given by:
[tex]P(X \geq 0) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = 0) = C_{84,0}.(\frac{1}{53})^{0}.(\frac{52}{53})^{84} = 0.2019[/tex]
[tex]P(X \geq 0) = 1 - P(X = 0) = 1 - 0.2019 = 0.7981[/tex]
79.81% probability of winning.
