A radio station runs a promotion at an auto show with a money box with $$ 13 \$ 50 $$ tickets, $$ 13 \$ 25 $$ tickets, and $$ 15 \$ 5 $$ tickets. The box contains an additional 20 "dummy" tickets with no value. Three tickets are randomly drawn. Find the probability that exactly two $$ \$ 50 $$ prizes and no other money winners are chosen. The probability that exactly two $$ \$ 50 $$ prizes and no other money winners are chosen is (Round to four decimal places as needed.)

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UpStudy AI · Accepted Answer

Let the total number of tickets be
$$
13 + 13 + 15 + 20 = 61.
$$
We are drawing 3 tickets from this set. We want exactly two $$\$50$$ tickets and no other money-winning tickets (i.e. the third ticket must be a "dummy" ticket with no value).

1. The number of ways to choose exactly two $$\$50$$ tickets from the 13 available is
   $$
   \binom{13}{2}.
   $$

2. The number of ways to choose one dummy ticket from the 20 available is
   $$
   \binom{20}{1}.
   $$

3. The total number of ways to choose any 3 tickets from 61 is
   $$
   \binom{61}{3}.
   $$

Thus, the probability is given by
$$
\text{Probability} = \frac{\binom{13}{2} \times \binom{20}{1}}{\binom{61}{3}}.
$$

Calculating each part:

- $$\binom{13}{2}$$ is
  $$
  \binom{13}{2} = \frac{13 \times 12}{2} = 78.
  $$
- $$\binom{20}{1} = 20.$$
- $$\binom{61}{3}$$ is
  $$
  \binom{61}{3} = \frac{61 \times 60 \times 59}{6} = 35,\!990.
  $$

Now, the probability becomes
$$
\frac{78 \times 20}{35,\!990} = \frac{1,\!560}{35,\!990}.
$$

Computing the value:
$$
\frac{1,\!560}{35,\!990} \approx 0.0434.
$$

Thus, the probability that exactly two $$\$50$$ prizes (and no other money winners) are drawn is approximately $$0.0434$$.
The probability of drawing exactly two \$50 tickets and no other money winners is approximately 0.0434.

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