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:
![Probability =\frac{Favourable \ outcome}{Total\ outcome}](https://tex.z-dn.net/?f=Probability%20%3D%5Cfrac%7BFavourable%20%5C%20outcome%7D%7BTotal%5C%20outcome%7D)
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 = ![^{15} C_{0}](https://tex.z-dn.net/?f=%5E%7B15%7D%20C_%7B0%7D)
Favorable outcome for drawing 3 dummy tickets out of 11 $25 tickets = ![^{11} C_{0}](https://tex.z-dn.net/?f=%5E%7B11%7D%20C_%7B0%7D)
Favorable outcome for drawing 3 dummy tickets out of 15 $5 tickets = ![^{15} C_{0}](https://tex.z-dn.net/?f=%5E%7B15%7D%20C_%7B0%7D)
Favorable outcome for drawing 3 dummy tickets out of 20 dummy ticket = ![^{20} C_{3}](https://tex.z-dn.net/?f=%5E%7B20%7D%20C_%7B3%7D)
Thus, total Favorable outcome for drawing 3 dummy tickets =
![^{15} C_{0}\times^{11} C_{0}\times^{15} C_{0}\times^{20} C_{3}](https://tex.z-dn.net/?f=%5E%7B15%7D%20C_%7B0%7D%5Ctimes%5E%7B11%7D%20C_%7B0%7D%5Ctimes%5E%7B15%7D%20C_%7B0%7D%5Ctimes%5E%7B20%7D%20C_%7B3%7D)
![\frac{15!}{(15-0)!0!} \times\frac{11!}{(11-0)!0!} \times\frac{15!}{(15-0)!0!} \times\frac{20!}{(20-3)!3!}](https://tex.z-dn.net/?f=%5Cfrac%7B15%21%7D%7B%2815-0%29%210%21%7D%20%5Ctimes%5Cfrac%7B11%21%7D%7B%2811-0%29%210%21%7D%20%5Ctimes%5Cfrac%7B15%21%7D%7B%2815-0%29%210%21%7D%20%5Ctimes%5Cfrac%7B20%21%7D%7B%2820-3%29%213%21%7D)
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B15%21%7D%7B15%21%5Ctimes1%7D%20%5Ctimes%5Cfrac%7B11%21%7D%7B11%21%5Ctimes1%7D%20%5Ctimes%5Cfrac%7B15%21%7D%7B15%21%5Ctimes1%7D%20%5Ctimes%5Cfrac%7B20%5Ctimes19%5Ctimes18%5Ctimes17%21%7D%7B17%21%5Cimes3%5Ctimes2%5Ctimes1%7D%20%5C%5C%5C%5C%201%5Ctimes1%5Ctimes1%5Ctimes%5Cfrac%7B6840%7D%7B6%7D%20%5C%5C%5C%5C%20%3D1140)
Total outcome for drawing 3 dummy tickets out of 61 tickets = ![^{61} C_{3}=\frac{61!}{(61-3)!\times3} =\frac{61\times60\times59\times58!}{58!\times3\times2\times}=\frac{215940}{6} =35990](https://tex.z-dn.net/?f=%5E%7B61%7D%20C_%7B3%7D%3D%5Cfrac%7B61%21%7D%7B%2861-3%29%21%5Ctimes3%7D%20%3D%5Cfrac%7B61%5Ctimes60%5Ctimes59%5Ctimes58%21%7D%7B58%21%5Ctimes3%5Ctimes2%5Ctimes%7D%3D%5Cfrac%7B215940%7D%7B6%7D%20%3D35990)
Now,
![Probability =\frac{Favourable \ outcome}{Total\ outcome}\\](https://tex.z-dn.net/?f=Probability%20%3D%5Cfrac%7BFavourable%20%5C%20outcome%7D%7BTotal%5C%20outcome%7D%5C%5C)
![=\frac{1140}{35990} =0.0316\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1140%7D%7B35990%7D%20%3D0.0316%5C%5C)
Thus, the probability of three tickets that are randomly drawn are dummy tickets is 0.0316