# Download Analytical Methods in Probability Theory: Proceedings by Dogeu D. (ed.), Lucaks E. (ed.), Rohatgi V.K. (ed.) PDF By Dogeu D. (ed.), Lucaks E. (ed.), Rohatgi V.K. (ed.)

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Example text

This sample space would not be simple because the outcomes would not be equally probable. 4 Genetics. Inherited traits in humans are determined by material in speciﬁc locations on chromosomes. Each normal human receives 23 chromosomes from each parent, and these chromosomes are naturally paired, with one chromosome in each pair coming from each parent. For the purposes of this text, it is safe to think of a gene as a portion of each chromosome in a pair. The genes, either one at a time or in combination, determine the inherited traits, such as blood type and hair color.

Proof It is known from Axiom 1 that Pr(A) ≥ 0. 4 implies Pr(A) ≤ Pr(S) = 1, by Axiom 2. 4. For every two events A and B, Pr(A ∩ B c ) = Pr(A) − Pr(A ∩ B). 11, the events A ∩ B c and A ∩ B are disjoint and A = (A ∩ B) ∪ (A ∩ B c ). 2 that Pr(A) = Pr(A ∩ B) + Pr(A ∩ B c ). Subtract Pr(A ∩ B) from both sides of this last equation to complete the proof. 7 For every two events A and B, Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B). 11, we have A ∪ B = B ∪ (A ∩ B c ), and the two events on the right side of this equation are disjoint.

10. 10, we sampled with replacement, but we distinguished between samples having the same balls in different orders. This could be called ordered sampling with replacement. 4, samples containing the same genes in different orders were considered the same outcome. This could be called unordered sampling with replacement. The general formula for the , number of unordered samples of size k with replacement from n elements is n+k−1 k and can be derived in Exercise 19. It is possible to have k larger than n when sampling with replacement.