A bicycle lock has a 4-digit combination. Each of the digits is a number from 0 to 9. Find the probability that the lock has a combination in which 3 of the 4 digits are 3s
Reiny was close, I would adjust the possible combination to 10^4 = 10,000
So 36/10,000 = 9/2,500
To find the probability, we need to determine the number of favorable outcomes (combinations with 3 digits as 3s) divided by the total number of possible outcomes.
Step 1: Determine the number of favorable outcomes:
There are 4 digits in the combination, and we want 3 of them to be 3s. The remaining digit can be any number from 0 to 9, excluding 3.
Since there are 10 possible digits (0-9) and one of them is already fixed as 3, we have 9 options for the remaining digit.
So, the number of favorable outcomes is 1 (fixed 3) multiplied by 9 (options for the remaining digit) = 9.
Step 2: Determine the total number of possible outcomes:
Each of the 4 digits in the combination can be any number from 0 to 9, so we have 10 options for each digit.
Therefore, the total number of possible outcomes is 10 (options for the first digit) multiplied by 10 (options for the second digit) multiplied by 10 (options for the third digit) multiplied by 10 (options for the fourth digit) = 10,000.
Step 3: Calculate the probability:
The probability of an event is given by the number of favorable outcomes divided by the total number of possible outcomes.
Therefore, the probability of the lock having a combination with 3 of the 4 digits being 3s is:
9 (favorable outcomes) divided by 10,000 (total possible outcomes) = 9/10,000.
Thus, the probability is 9/10,000 or 0.0009 (or 0.09%)
To find the probability that the lock has a combination in which 3 of the 4 digits are 3s, we need to consider the total number of possible combinations and the number of combinations that satisfy the given condition.
The total number of possible combinations can be obtained by multiplying the number of options for each digit. Since each digit can be any number from 0 to 9, there are 10 options for each digit. Therefore, the total number of possible combinations is 10 * 10 * 10 * 10 = 10,000.
Now let's find the number of combinations that satisfy the given condition, where 3 of the 4 digits are 3s.
We have four positions for the digits in the combination: _ _ _ _
The first digit can be any number from 0 to 9, so we have 10 options.
The second digit can also be any number from 0 to 9, including 3 because we need at least 3 digits to be 3. So again, we have 10 options.
The third digit must be 3 since we want exactly 3 of the 4 digits to be 3. So we have only 1 option, which is 3.
The fourth digit must also be 3, again giving us 1 option.
Therefore, the number of combinations that satisfy the given condition is 10 * 10 * 1 * 1 = 100.
Now, we can find the probability by dividing the favorable outcomes (the number of combinations satisfying the condition) by the total possible outcomes.
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 100 / 10,000
Probability = 0.01 or 1%
Therefore, the probability that the lock has a combination in which 3 of the 4 digits are 3s is 0.01 or 1%.
I will assume that repeats are allowed as in most combination locks
so number of possible codes = 10^3 = 1000
So we want three 3's and one non-3
333x
33x3
3x33
x333---> where x is a number other than 3
so there are 4(9) of these or 36
so prob of our event = 36/1000 = 9/250