You play a game that involves drawing two numbers from a hat. There are 25 pieces of paper numbered from 1 to 25 in the hat. Each number is replaced after it is drawn. Find the probability that you will draw the 3 on your first draw and a number greater than 10 on your second draw. Round your answer to the nearest tenth.about $\%$

To solve this problem, we need to find the probability of two independent events happening in sequence.

1. Probability of drawing a 3 on the first draw:

   • There is only one piece of paper with the number 3 out of 25 pieces of paper.
   • Therefore, the probability of drawing a 3 on the first draw is: $P(\text{First draw is 3}) = \frac{1}{25}$

2. Probability of drawing a number greater than 10 on the second draw:

   • Numbers greater than 10 are 11, 12, 13, ..., 25. There are 15 such numbers.
   • Since the number is replaced after the first draw, the total number of pieces of paper remains 25.
   • Therefore, the probability of drawing a number greater than 10 on the second draw is: $P(\text{Second draw is greater than 10}) = \frac{15}{25} = \frac{3}{5}$

3. Combined probability of both events happening:

   • Since the two events are independent, we multiply their probabilities: $P(\text{First draw is 3 and second draw is greater than 10}) = P(\text{First draw is 3}) \times P(\text{Second draw is greater than 10})$ $= \frac{1}{25} \times \frac{3}{5} = \frac{1 \times 3}{25 \times 5} = \frac{3}{125}$

4. Convert the probability to a percentage:

   • To convert the fraction to a percentage, we multiply by 100: $\frac{3}{125} \times 100 = 2.4\%$

Therefore, the probability that you will draw the 3 on your first draw and a number greater than 10 on your second draw is approximately 2.4%.