1.4.2 Sampling with Replacement Flashcards
(13 cards)
What does ‘sampling with replacement’ mean?
After selecting an item, you put it back into the pool before the next selection. Probabilities stay constant for each draw.
Why is ‘sampling with replacement’ easier to compute in many cases?
Because the pool doesn’t change, you can multiply the same number of choices across positions.
What is the key difference between sampling with and without replacement?
With replacement, probabilities don’t change. Without replacement, the sample space shrinks each time.
In sampling with replacement, how do you generally solve problems?
Use multiplication and basic logic; there’s rarely a general formula.
How many 3-letter codes can be made using only letters with replacement?
26 × 26 × 26 = 17,576
Why is the answer to the 3-letter code question not 26 × 25 × 24?
That would be without replacement. With replacement, each letter slot still has 26 options.
How many 3-digit odd numbers can be made if the first digit cannot be 0?
9 × 10 × 5 = 450
Why is the first digit limited to 9 choices in the odd number example?
It cannot be 0, so only digits 1–9 are allowed.
Why is the last digit in the 3-digit number limited to 5 options?
Because it must be odd: 1, 3, 5, 7, or 9.
What is the alternative way to solve the 3-digit odd number problem?
Subtract 100 invalid numbers (000–099) and take half of the remaining 900 (since half will be odd): 900/2 = 450.
What is the arrangement count for 8 letters, 3 digits, 1 special character?
C(12,3,1,8) = 12! / (3! × 1! × 8!)
Why is it important to split the 12- and 13-character password calculations?
Because they have different structures and arrangement counts.
What makes sampling with replacement easier than without in terms of logic?
The number of options for each selection stays the same, so you can just multiply across stages.
