Respuesta :
Using the hypergeometric distribution, it is found that there is a 0.0333 = 3.33% probability that all three tickets have no value.
The tickets are drawn without replacement, hence, the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- There is a total of 15 + 13 + 12 + 20 = 60 tickets, hence [tex]N = 60[/tex].
- Of those, 20 are "dummy" tickets, hence [tex]k = 20[/tex].
- Three tickets are randomly drawn, hence [tex]n = 3[/tex].
The probability that all three tickets have no value is P(X = 0), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 0) = h(0,60,3,20) = \frac{C_{20,3}C_{40,0}}{C_{60,3}} = 0.0333[/tex]
0.0333 = 3.33% probability that all three tickets have no value.
A similar problem is given at https://brainly.com/question/24826394
