# Thus if f x is the pdf of x which we denote by

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Thus , if f ( x ) is the p.d.f. of x , which we denote by writing x f ( x ), then F ( x ) = x −∞ f ( x ) dx = P ( −∞ < x x ) . In the case where x is a discrete random variable with a probability mass function f ( x ), also denoted by x f ( x ), there is F ( x ) = x x f ( x ) = P ( −∞ < x x ) . Pascal’s Triangle and the Binomial Expansion. The binomial distribution is one of the fundamental distributions of mathematical statistics. To derive the bino- mial probability mass function from first principles, it is necessary to invoke some fundamental principles of combinatorial calculus. We shall develop the necessary results at some length in the ensuing sections before deriving the mass function. Consider the following binomial expansions: ( p + q ) 0 = 1 , ( p + q ) 1 = p + q, ( p + q ) 2 = p 2 + 2 pq + q 2 , 2

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( p + q ) 3 = p 3 + 3 p 2 q + 3 pq 2 + q 3 , ( p + q ) 4 = p 4 + 4 p 3 q + 6 p 2 q 2 + 4 pq 3 + q 4 , ( p + q ) 5 = p 5 + 5 p 4 q + 10 p 3 q 2 + 10 p 2 q 3 + 5 pq 4 + q 5 . The generic expansion is in the form of ( p + q ) n = p n + np n 1 q + n ( n 1) 2 p n 2 q 2 + n ( n 1)( n 2) 3! p n 3 q 3 + · · · + n ( n 1) · · · ( n r + 1) r ! p n r q r + · · · + n ( n 1)( n 2) 3! p 3 q n 3 + n ( n 1) 2 p 2 q n 2 + npq n 1 + p n . In a tidier notation, this becomes ( p + q ) n = n x =0 n ! ( n x )! x ! p x q n x . We can find the coeﬃcient of the binomial expansions of successive degrees by the simple device known as Pascal’s triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 The numbers in each row but the first are obtained by adding two adjacent numbers in the row above. The rule is true even for the units that border the triangle if we suppose that there are some invisible zeros extending indefinitely on either side of each row. Instead of relying merely upon observation to establish the formula for the binomial expansion, we should prefer to derive the formula by algebraic methods. Before we do so, we must reaﬃrm some notions concerning permutations and com- binations that are essential to a proper derivation. Permutations. Let us consider a set of three letters { a, b, c } and let us find the number of ways in which they can be can arranged in a distinct order. We may pick any one of the three to put in the first position. Either of the two remaining letters may be placed in the second position. The third position must be filled by the unused letter. With three ways of filling the first place, two of filling the second 3
and only one way of filling the third, there are altogether 3 × 2 × 1 = 6 different arrangements. These arrangements or permutations are abc, cab, bca, cba, bac acb. Now let us consider an unordered set of n objects denoted by { x i ; i = 1 , . . . , n } , and let us ascertain how many different permutations arise in this case. The answer can be found through a litany of questions and answers which we may denote by [( Q i , A i ); i = 1 , . . . , n ]: Q 1 : In how may ways can the first place be filled? A 1 : n Q 2 : In how may ways can the second place be filled? A 2 : n 1 ways, Q 3 : In how may ways can the third place be filled? A 3 : n 2 ways, .

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• Spring '12
• D.S.G.Pollock
• Probability theory, Probability mass function, Fill

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