Question

Jason bought 10 of the 30 raffle tickets for a drawing. Assuming that he chooses the winning ticket each time, what is the probability that Jason will win all 3 of the prizes if once a raffle ticket wins a prize, the ticket is thrown away?
StartFraction 6 over 203 EndFraction
StartFraction 1 over 27 EndFraction
StartFraction 3 over 29 EndFraction
StartFraction 1 over 9 EndFraction

Answer 1

To determine the probability that Jason will win all 3 prizes, we can analyze the problem step by step.

1. **Total Tickets:** There are 30 raffle tickets in total.
2. **Tickets Bought by Jason:** Jason bought 10 tickets.

When calculating probabilities, we need to consider that once a ticket wins a prize, it is not returned to the draw.

### Step 1: Probability of Winning the First Prize

The probability that Jason wins the first prize is the ratio of his tickets to the total tickets:
\[
P(\text{Jason wins 1st}) = \frac{10}{30} = \frac{1}{3}
\]

### Step 2: Probability of Winning the Second Prize

If Jason wins the first prize, there are now 29 tickets left (30 - 1), and Jason still has 9 winning tickets (10 - 1):
\[
P(\text{Jason wins 2nd}) = \frac{9}{29}
\]

### Step 3: Probability of Winning the Third Prize

If Jason wins the second prize, there are now 28 tickets left (29 - 1), and Jason still has 8 winning tickets (9 - 1):
\[
P(\text{Jason wins 3rd}) = \frac{8}{28} = \frac{2}{7}
\]

### Step 4: Overall Probability

To find the total probability that Jason wins all three prizes, we multiply the individual probabilities:
\[
P(\text{All Wins}) = P(\text{Jason wins 1st}) \times P(\text{Jason wins 2nd}) \times P(\text{Jason wins 3rd})
\]

Calculating this:
\[
P(\text{All Wins}) = \left(\frac{1}{3}\right) \times \left(\frac{9}{29}\right) \times \left(\frac{2}{7}\right)
\]

Now simplify it:
1. Multiply the numerators:
\[
1 \times 9 \times 2 = 18
\]

2. Multiply the denominators:
\[
3 \times 29 \times 7 = 609
\]

3. So, the probability that Jason wins all 3 prizes is:
\[
P(\text{All Wins}) = \frac{18}{609}
\]

### Step 5: Simplifying the Fraction

Both the numerator and the denominator can be divided by 3:
\[
\frac{18 \div 3}{609 \div 3} = \frac{6}{203}
\]

Thus, the probability that Jason will win all 3 prizes is:
\[
\boxed{\frac{6}{203}}
\]