A six-sided die has been tossed 4 times. How many sequences of these 4 outcomes contain at least one number greater than 4?
1040
Well, there are a total of 6^4 = 1296 possible sequences when tossing a six-sided die 4 times. Now, let's figure out how many sequences do not contain any number greater than 4.
If we think about it, there are 4 choices for each toss (1, 2, 3, or 4), so there are 4^4 = 256 sequences in which no number is greater than 4.
Therefore, the number of sequences that contain at least one number greater than 4 is 1296 - 256 = 1040.
So, there are 1040 sequences that will make you feel lucky and have you saying, "Rolling high numbers, four times more fun!"
To find the number of sequences of 4 outcomes where at least one number is greater than 4, you can calculate the total number of sequences of 4 outcomes and subtract the number of sequences where all the numbers are less than or equal to 4.
Step 1: Calculate the total number of sequences of 4 outcomes.
For each toss of the die, there are 6 possible outcomes (numbers 1 to 6). Since the die has been tossed 4 times, you can multiply the number of outcomes at each toss:
6 * 6 * 6 * 6 = 6^4 = 1,296
Step 2: Calculate the number of sequences where all the numbers are less than or equal to 4.
Since all the numbers must be less than or equal to 4, there are 4 possible outcomes at each toss (numbers 1 to 4). Hence, the number of sequences where all the numbers are less than or equal to 4 is:
4 * 4 * 4 * 4 = 4^4 = 256
Step 3: Calculate the number of sequences where at least one number is greater than 4.
To calculate this, you need to subtract the number of sequences where all the numbers are less than or equal to 4 from the total number of sequences:
1,296 - 256 = 1,040
Therefore, there are 1,040 sequences of 4 outcomes that contain at least one number greater than 4.