Final answer:
The probability of needing at most 3 rolls to get a 4 is 0.5787. The probability of needing an even number of die rolls is 0.5.
Step-by-step explanation:
To find the probability that we need at most 3 rolls to get a 4, we can consider the possible outcomes for each roll. On the first roll, there are 6 possible outcomes (numbers 1-6). On the second roll, there are also 6 possible outcomes. Similarly, on the third roll, there are 6 possible outcomes. Therefore, the total number of possible outcomes for 3 rolls is 6 * 6 * 6 = 216.
To determine the desired outcomes, we need to consider that we stop rolling once a 4 appears. There are 5 possible outcomes for each roll (numbers 1, 2, 3, 5, and 6). Therefore, the total number of desired outcomes for 3 rolls is 5 * 5 * 5 = 125.
The probability that we need at most 3 rolls is the ratio of desired outcomes to possible outcomes: 125/216 = 0.5787 (rounded to four decimal places).
To find the probability that we needed an even number of die rolls, we need to consider the possible combinations of even and odd rolls. The first roll can be either even or odd, so there are 2 possible outcomes. For each subsequent roll, the number of possible outcomes remains the same. Therefore, the total number of possible outcomes for any number of rolls is 2 * 2 * 2 * ... (n times), where n is the number of rolls.
Since we stop rolling once a 4 appears, the last roll must be even. Therefore, the number of desired outcomes for an even number of rolls is 2 * 2 * 2 * ... (n-1 times), where n is an even number.