The serial number on a $10 bill contains eight non-zero digits. If your $10 bill contains the digit 7 at least once, you win a prize. The probability that your $10 bill will be a prize-winning bill is what?
I believe I answered this in an earlier post.
There are 9 possible values for the first digit and 9 for the second, and so on. So, there are 9^8 possible serial numbers, where no digits is a zero. Similarly, there are 8^8 possible values where no digit is a zero or a seven. So, the probability there is at least one seven is:
1- (8^8)/(9^8)
To find the probability of winning a prize with your $10 bill, we first need to determine the total number of possible combinations of eight non-zero digits.
Since the bill contains eight non-zero digits, each digit can be any number from 1 to 9 (excluding 0). Therefore, we have 9 options for each digit.
Hence, the total number of possible combinations is 9^8 (9 raised to the power of 8).
Next, we need to determine the number of combinations that include the digit 7 at least once. To calculate this, we can make use of the complement principle.
The complement principle states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring.
So, in this case, let's calculate the probability of the event of not winning, which means not having the digit 7 in the serial number.
The number of combinations without the digit 7 can be determined by considering that we have 8 positions to be filled, and each position has 8 options (any digit from 1 to 9, excluding 7). Therefore, the number of combinations without 7 is 8^8.
Now, let's determine the probability of not winning: P(not winning) = 8^8 / 9^8
Finally, we can determine the probability of winning by using the complement principle:
P(winning) = 1 - P(not winning)
P(winning) = 1 - (8^8 / 9^8)
By calculating this expression, we can find the probability that your $10 bill will be a prize-winning bill.