Respuesta :
Answer:
0.0316
Step-by-step explanation:
Given:
A radio station runs a promotion at an auto show with a money box with 15 $50 tickets, 11 $25 tickets, and 15 $5 tickets.
The box contains an additional 20 "dummy" tickets with no value.
Thus, total number of tickets = 15 + 11 + 15 + 20 = 61
Three tickets are randomly drawn.
Question asked:
Find the probability that all three tickets have no value.
Solution
As we know:
[tex]Probability =\frac{Favourable \ outcome}{Total\ outcome}[/tex]
First of all we will find favorable outcome for drawing 3 dummy tickets.
Favorable outcome for drawing 3 dummy tickets out of 15 $50 tickets = [tex]^{15} C_{0}[/tex]
Favorable outcome for drawing 3 dummy tickets out of 11 $25 tickets = [tex]^{11} C_{0}[/tex]
Favorable outcome for drawing 3 dummy tickets out of 15 $5 tickets = [tex]^{15} C_{0}[/tex]
Favorable outcome for drawing 3 dummy tickets out of 20 dummy ticket = [tex]^{20} C_{3}[/tex]
Thus, total Favorable outcome for drawing 3 dummy tickets =
[tex]^{15} C_{0}\times^{11} C_{0}\times^{15} C_{0}\times^{20} C_{3}[/tex]
[tex]\frac{15!}{(15-0)!0!} \times\frac{11!}{(11-0)!0!} \times\frac{15!}{(15-0)!0!} \times\frac{20!}{(20-3)!3!}[/tex]
[tex]\frac{15!}{15!\times1} \times\frac{11!}{11!\times1} \times\frac{15!}{15!\times1} \times\frac{20\times19\times18\times17!}{17!\imes3\times2\times1} \\\\ 1\times1\times1\times\frac{6840}{6} \\\\ =1140[/tex]
Total outcome for drawing 3 dummy tickets out of 61 tickets = [tex]^{61} C_{3}=\frac{61!}{(61-3)!\times3} =\frac{61\times60\times59\times58!}{58!\times3\times2\times}=\frac{215940}{6} =35990[/tex]
Now,
[tex]Probability =\frac{Favourable \ outcome}{Total\ outcome}\\[/tex]
[tex]=\frac{1140}{35990} =0.0316\\[/tex]
Thus, the probability of three tickets that are randomly drawn are dummy tickets is 0.0316
