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Sagot :
Let's solve the problem in a step-by-step manner.
We start with a raffle where Jason has bought 10 out of the 30 tickets, and we need to determine the probability that Jason will win all three prizes, given that each ticket is removed after winning a prize.
1. Probability of Winning the First Prize:
- Jason has 10 out of 30 tickets.
- The probability that Jason wins the first prize is the ratio of Jason's tickets to the total tickets.
[tex]\[ \text{Probability of winning the first prize} = \frac{10}{30} = \frac{1}{3} \][/tex]
2. Probability of Winning the Second Prize:
- If Jason wins the first prize, one of his tickets is removed, leaving him with 9 tickets.
- The total number of tickets now becomes 29.
- The probability that Jason wins the second prize is the ratio of Jason's remaining tickets to the remaining total tickets.
[tex]\[ \text{Probability of winning the second prize} = \frac{9}{29} \][/tex]
3. Probability of Winning the Third Prize:
- If Jason wins the second prize as well, another one of his tickets is removed, leaving him with 8 tickets.
- The total number of tickets now becomes 28.
- The probability that Jason wins the third prize is the ratio of Jason's remaining tickets to the remaining total tickets.
[tex]\[ \text{Probability of winning the third prize} = \frac{8}{28} = \frac{2}{7} \][/tex]
4. Total Probability of Winning All Three Prizes:
- The total probability of Jason winning all three prizes is the product of the probabilities of winning each prize.
[tex]\[ \text{Total Probability} = \left(\frac{1}{3}\right) \times \left(\frac{9}{29}\right) \times \left(\frac{2}{7}\right) \][/tex]
By multiplying these individual probabilities, we get:
[tex]\[ \frac{1}{3} \times \frac{9}{29} \times \frac{2}{7} \approx 0.333 \times 0.310 \times 0.286 \approx 0.02956 \][/tex]
So, the probability that Jason will win all three prizes is approximately [tex]\(0.02956\)[/tex].
Hence, this probability, when simplified to a fraction, is not immediately obvious among the provided options:
[tex]\[ \frac{6}{203}, \frac{1}{27}, \frac{3}{29}, \frac{1}{9} \][/tex]
Upon further examination and cross-verifying with approximate decimal values:
[tex]\[ 0.02956 \approx \frac{6}{203} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{\frac{6}{203}} \][/tex]
We start with a raffle where Jason has bought 10 out of the 30 tickets, and we need to determine the probability that Jason will win all three prizes, given that each ticket is removed after winning a prize.
1. Probability of Winning the First Prize:
- Jason has 10 out of 30 tickets.
- The probability that Jason wins the first prize is the ratio of Jason's tickets to the total tickets.
[tex]\[ \text{Probability of winning the first prize} = \frac{10}{30} = \frac{1}{3} \][/tex]
2. Probability of Winning the Second Prize:
- If Jason wins the first prize, one of his tickets is removed, leaving him with 9 tickets.
- The total number of tickets now becomes 29.
- The probability that Jason wins the second prize is the ratio of Jason's remaining tickets to the remaining total tickets.
[tex]\[ \text{Probability of winning the second prize} = \frac{9}{29} \][/tex]
3. Probability of Winning the Third Prize:
- If Jason wins the second prize as well, another one of his tickets is removed, leaving him with 8 tickets.
- The total number of tickets now becomes 28.
- The probability that Jason wins the third prize is the ratio of Jason's remaining tickets to the remaining total tickets.
[tex]\[ \text{Probability of winning the third prize} = \frac{8}{28} = \frac{2}{7} \][/tex]
4. Total Probability of Winning All Three Prizes:
- The total probability of Jason winning all three prizes is the product of the probabilities of winning each prize.
[tex]\[ \text{Total Probability} = \left(\frac{1}{3}\right) \times \left(\frac{9}{29}\right) \times \left(\frac{2}{7}\right) \][/tex]
By multiplying these individual probabilities, we get:
[tex]\[ \frac{1}{3} \times \frac{9}{29} \times \frac{2}{7} \approx 0.333 \times 0.310 \times 0.286 \approx 0.02956 \][/tex]
So, the probability that Jason will win all three prizes is approximately [tex]\(0.02956\)[/tex].
Hence, this probability, when simplified to a fraction, is not immediately obvious among the provided options:
[tex]\[ \frac{6}{203}, \frac{1}{27}, \frac{3}{29}, \frac{1}{9} \][/tex]
Upon further examination and cross-verifying with approximate decimal values:
[tex]\[ 0.02956 \approx \frac{6}{203} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{\frac{6}{203}} \][/tex]
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