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Summary What have we learned?.

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Presentation on theme: "Summary What have we learned?."— Presentation transcript:

1 Summary What have we learned?

2 Coin Toss Combine group experimental results Theoretical results
40 limit/or extra tosses to have exactly 96 30 Theoretical results mult. by? 20 0H, 5T ___ ___ 1H, 4T ___ ___ 2H, 3T ___ ___ 3H, 2T ___ ___ 4H, 1T ___ ___ 5H, 0T ___ ___ Total ___ ___ 10 96 0H 1H 2H 3H 4H 5H How do the theoretical and experimental results compare?

3 Pascal’s Triangle 1 nC0 nC1 nC nCr nCn-1 nCn row 0 row 1 row 2 row 3 row 4 row 5 . . row n nCr “n choose r” the number of ways you can get r objects when selecting from n objects “object” can be a thing or an event

4 Binomial Theorem (x + y) two objects, x and y, one time
compare to H or T (x + y)2 = (x + y)(x + y) two objects, x and y, two times when multiply by x and then by y, end up with all possible combinations of the two variables multiplied together compare to HH,HT,TH, TT (x + y)3 = (x + y)(x + y)(x + y) two objects, three times all possible combinations of the two variables multiplied together three times compare to ,HHH,HHT,HTH,HTT,THH,THT,TTH, TTT Pascal’s Triangle gives the coefficients (how many times each group occurs), then append the variables to each term (x + y)n = nC0xn + nC1xn-1y + nC2xn-2y + nCrxn-ryr nCn-1x1yn-1 + nCnyn x exponent is number of x’s, y exponent is r, the number of y’s sum of the exponents always equals n

5 n(n-1)(n-2)…2·1 = n! Permutations of n objects
nCr How many ways can we order n objects? n(n-1)(n-2)…2·1 = n! Permutations of n objects What if we are interested in r of the n objects? (but order still matters) 1st, 2nd, and 3rd finishers in a race with 9 contestants 9P3 = 9·8·7 n(n-1)(n-2)…2· n! Permutations of n objects taken r at a time (n-r)…2· (n-r)! nPr ways we can have the “don’t care” combinations, the ordering of the n-r remaining objects What if we are interested in r of the n objects? (and order doesn’t matter) n(n-1)(n-2)…2· n! Combinations of n objects taken r at a time (n-r)…2·1(r)(r-1)… 2·1 (n-r)!r! nCr = r!, the number of ways we can order the r objects


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