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

1 answer

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: \[ \boxed{\frac{6}{203}} \]