g a ticket between 3 and 9 in a single draw from the described box is 50%, as only the tickets marked 9 fall into this range. For multiple draws with replacement, each draw remains independent with the same probability. However, without additional information on how to approximate over 100 draws, a precise answer isn't provided here.
The student's question involves calculating the approximate chance of drawing a ticket with a number between 3 and 9 (inclusive) from a box when drawing 100 tickets with replacement. The box contains one ticket marked 1, four tickets marked 2, and five tickets marked 9. To find the probability of drawing a ticket between 3 and 9, we only need to consider the tickets marked 9 since those are the only ones within the desired range.
The probability of drawing a 9 on any single draw is the number of 9s (5) over the total tickets (10). Therefore, the chance of drawing a 9 in one trial is 0.5. Since each draw is independent, and because the draws are with replacement, this probability remains constant across all 100 draws.
For calculating the approximate chance over 100 draws, we would typically use the binomial distribution. However, for a large number of trials, this becomes impractical. Instead, one might approximate the distribution of the sum of a large number of independent binary variables using the normal distribution (central limit theorem). The number of successful draws (drawing a 9) in 100 trials would then be approximated using this method. Yet, a full calculation of the approximate chances over 100 trials using the central limit theorem is beyond the scope of this platform and the level of the original question.
In conclusion, the chance of drawing a ticket between 3 and 9 in a single draw is 0.5 or 50%. Because the question asks for an approximate chance over 100 draws and specifics of the approximation method are not provided, we acknowledge that the problem is complex and not completely addressed in this response.