Week 17 (Bonus Probability) Flashcards
An urn contains n balls labelled 1, 2, β¦, n.
Suppose I draw r balls randomly from the urn, replacing each ball before I draw the next one.
What is the probability that all r balls have different numbers?
(Think about this systematically to derive the formula. There are 5 simple steps.)
Remember that the first draw has no effect on the probability calculation here.
- For the second drawn ball to be different, the probability is (π-1)/π, e.g. if 25 uniquely-labelled balls, 24 of them will be different to the first one drawn
- Repeat this pattern for each subsequent draw, noticing that the last draw is the (π-1)th one because the first one is excluded from the probability calculation:
(π-2)/n, (π-3)/π, β¦, (π-[π-1])/π
When we multiply all of these terms together, we have an expression like [(π-1)(π-2)β¦(π-[r-1])]/ππ-1 - Multiply by π/π to make the denominator ππ and prepend π to the numerator
- Probably factor 1/ππ to make it easier to read for a sec - then multiply the remaining term by [(π-π)!/(π-π)!] - but all written out as products - because then the numerator ends up being π! while the denominator remains (π-π)!
- Beautiful! Donβt forget that factored-out 1/ππ - then youβve got your formula:
π! / ππ(π-π)! β¨
An urn contains n balls labelled 1, 2, β¦, n.
Suppose I draw r balls randomly from the urn, replacing each ball before I draw the next one.
What is the probability that among the balls I draw, I can find two with the same number?
What is the important thing to remember about this question?
Important to remember that this means at least two balls were drawn with the same number. Not exactly two!
Find the number 3 letter words that can be formed from the letters a, b, c, d, and e in which the letters are allowed to be repeated.
The number of letters available is n = 5.
The number of letters in each word is r = 3.
Permutations with repetition: nr
Since there can be the repetition of letters, the possible number of words is, 53 = 125.
