Based on the experiment illustrated in the picture and the experiment that I conducted, I concluded to the following results. The probability of getting a possible 3 if we roll one die, is 1/6 as the die has 6 sides and the number 3 is only in 1 side of the die. Although, if we roll two dice, the probability of getting a possible 3 in at least one die, is twice as likely to occur with a chance of 1/6 in the one die plus 1/6 on the other, which is equal to 2/6 for both dice and therefore simplified to 1/3. In the meantime, there are six conditions that the die A might get a 3 and six conditions that die B might be the same number as die A; thus one condition in both dice that they get a possible 3. The probability for the example explained is 12/36 minus 1/36, which is equal to 11/36. …show more content…
The multiplication principle is obtained by 6 possibilities for both dice, consequently having an outcome of 6 times 6 is equaled to 36. On the other hand, the probability of one dice not being a possible number for example number 3 is 5/6. If I don’t want to roll number 3 in both dice I will multiply 5/6 with 5/6 and get a result of 25/36. A mentioned before, the probability of someone not getting a 3 is 1/6, so the probability of two dice being the possible number 3 is 1/6 times 1/6 that is equal to 1/36. This is a different concept as saying that both dice have rolled a possible 3 because there are six different possible numbers, which they would be 6/36 or 1/6. Moreover, the opposite of both dice that have the number 3, the probabilities will sum up to one. But the probability of at least one die not being the number 3, equaling to 1 minus 1/36 that is equal to 35/36. A simpler way to acknowledge this condition is to think of it as the opposite of the next example and situation, where the two dice are not either number, for example a possible 3 and a possible