Answer:
the probability of neither showing a multiple of 3 nor a multiple of 4 on the second die is 17/36, or approximately 0.4722
Explanation:
To find the probability of neither showing a multiple of 3 nor a multiple of 4 on the second die, we need to first determine the sample space of possible outcomes.
The sample space of rolling two dice is given by all possible pairs of numbers from 1 to 6, which has a total of 6 x 6 = 36 possible outcomes.
Out of these 36 possible outcomes, there are 12 outcomes where the second die shows a multiple of 3 (namely 3, 6, and 9), and 9 outcomes where the second die shows a multiple of 4 (namely 4 and 8).
However, we must note that there are two outcomes where the second die shows both a multiple of 3 and a multiple of 4, namely (3,4) and (6,4). We need to subtract these two outcomes from our total count.
Therefore, the number of outcomes where the second die shows a multiple of 3 or a multiple of 4 or both is 12 + 9 - 2 = 19.
The number of outcomes where the second die does not show a multiple of 3 or a multiple of 4 is the complement of the above count, which is 36 - 19 = 17.
Therefore, the probability of neither showing a multiple of 3 nor a multiple of 4 on the second die is 17/36, or approximately 0.4722.