Suppose that a 4-digit pin number is to be formed with the digits in {0, 1, ,2 ,3 ,4 ,5, 6, 7, 8, 9}. What is the probability that the pin number contains at least 3 odd digits
To find the probability of a pin number containing at least 3 odd digits, we need to determine the total number of possible pin numbers and the number of pin numbers that meet our condition.
Step 1: Calculate the total number of possible pin numbers.
Since there are 10 digits to choose from (0-9) for each of the 4 positions, the total number of possible pin numbers is 10 × 10 × 10 × 10 = 10,000.
Step 2: Calculate the number of pin numbers meeting our condition.
To have at least 3 odd digits, we can have either 3 or 4 odd digits.
Case 1: 3 odd digits
Selecting 3 odd digits can be done in C(5,3) ways, where C(n,r) denotes the combination of choosing r items from a total of n items. In this case, we have 5 odd digits: {1, 3, 5, 7, 9}. Thus, C(5,3) = 10.
For the remaining 1 or 2 even digits, we have 5 options (0, 2, 4, 6, 8) to choose from for each position, giving us 5 × 5 × 5 = 125 possibilities.
Therefore, the total number of pin numbers with 3 odd digits is 10 × 125 = 1,250.
Case 2: 4 odd digits
The case of having 4 odd digits can be solved similarly.
C(5,4) = 5 ways to choose 4 odd digits from {1, 3, 5, 7, 9}.
For the remaining 0 or 1 even digits, we again have 5 options for each position, resulting in 5 × 5 × 5 = 125 possibilities.
Thus, the total number of pin numbers with 4 odd digits is 5 × 125 = 625.
Step 3: Calculate the probability.
To find the probability, we divide the number of pin numbers meeting our condition by the total number of possible pin numbers:
Probability = (Number of pin numbers meeting the condition) / (Total number of possible pin numbers)
Probability = (1,250 + 625) / 10,000
Probability = 1,875 / 10,000
Probability = 0.1875
Therefore, the probability that a pin number contains at least 3 odd digits is 0.1875 or 18.75%.