Supp#3COMBINATORICS_AND_PROBABILITY_DISTRIBUTIONS

Supp#3COMBINATORICS_AND_PROBABILITY_DISTRIBUTIONS - EE 150...

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1 EE 150 COMBINATORICS AND PROBABILITY DISTRIBUTIONS We have shown that the number of subsets of k objects that can be wrought from an original group of n objects [the order in which the objects are placed in each subset being unimportant] is: C(n,k) = [Equation one] This formula presents itself in many situations. Imagine that you toss a coin five times and you want the number of ways that you can toss two heads in five tosses. Tossing a head on the first and fourth toss represents a unique subset of two taken out of an original group of five. So the number of ways of tossing two heads in five tosses is simple C(5,2) . Remarkably if you expand a binomial to some power the coefficients in the expansion are given by the combination formula in particular: n ( x+y) n = Σ [ n ! / k ( n – k ) ] x k y n - k k=0 [Equation two]
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2 Examples of the expansion written out completely shows the pattern of the coefficients: [Equations Three] If you plot the coefficients of the binomial expansion you will get a bell shaped or gaussian curve. As n gets larger and larger the curve will get smoother and smoother. Let’s look at the expansion for the binomial to the fifth power. The first coefficient (one) is the number of ways that you can toss a coin five times and get five heads. The second coefficient (five) is the number of ways you can toss a coin five times and get four heads . The third coefficient (ten) is the number of ways you can toss a coin five times and get three heads. The fourth coefficient (ten) is the number of ways you can toss a coin five times and get two heads. The fifth coefficient (five) is the number of ways you can toss a coin five times and get one head. The sixth coefficient (one) is the number of ways you can toss a coin five times and get no heads. If you want the probability of say tossing four heads in five tosses that would be the the number of ways of tossing four heads divided by the number of possible outcomes. The number of possible outcomes for tossing a coin twice is 2 2 = 4 . The number of possible outcomes for tossing a coin three times is 2 3 = 8 . The number of possible outcomes for tossing a coin four times is 2 4 = 16 . The number of possible outcomes for tossing a coin five times is 2 5 = 32 . You can prove this to yourself by drawing the corresponding probability tree diagrams and counting. So the probability of tossing four heads ( and one tail ) in five tosses is 5/ 2 5 = 5/32 . Notice that in the expansion for the binomial expanded to the fifth power that if x is the probability of tossing heads on any given toss ( ½ for a fair coin ) and y is the probability of tossing tails on any given toss ( ½ for a fair coin ) , the second term in the expansion is 5 (1/2) 4 (1/2) = 5/ 2 5 = 5/32 , which is just simply the probability of tossing four heads in five tosses. The other complete terms in the expansion are the expected corresponding probabilities.
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Supp#3COMBINATORICS_AND_PROBABILITY_DISTRIBUTIONS - EE 150...

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