Question

A radio station runs a promotion at an auto show with a money box with 15 $100 tickets, 13 $50 tickets, and 12 $25 tickets. The box contains an additional 20 "dummy”tickets with no value. Three tickets are randomly drawn. Find the probability that all three tickets have no value.

Answer 1

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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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 $N = 60$.
• Of those, 20 are "dummy" tickets, hence $k = 20$.
• Three tickets are randomly drawn, hence $n = 3$.

The probability that all three tickets have no value is P(X = 0), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,60,3,20) = \frac{C_{20,3}C_{40,0}}{C_{60,3}} = 0.0333$

0.0333 = 3.33% probability that all three tickets have no value.

A similar problem is given at https://brainly.com/question/24826394