Probability - Robert P. Dobrow

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than Bs, at some point in the count, there must be a tie and the path crosses the -axis. Reflecting the portion of the path up until the tie across the -axis gives a “bad” path that starts with A.Having established a one-to-one correspondence we see that the number of “bad” lists that start with A is equal to the number of lists that start with B. There are lists that start with B. This gives the desired probabilityWe leave it to the reader to check that this last expression simplifies to

      Consider a sequence of events upper A 1 comma upper A 2 comma ellipsis period In this section, we consider strategies to find the probability that at least one of the events occurs, which is the probability of the union union upper A Subscript i, i.e., upper A 1 union upper A 2 union upper A 3 union midline-horizontal-ellipsis.

      Sometimes the complement of an event can be easier to work with than the event itself. The complement of the event that at least one of the upper A Subscript is occurs is the event that none of the upper A Subscript is occur, which is the intersection intersection upper A Subscript i Superscript c Baseline period

      Check with a Venn diagram (and if you are comfortable working with sets prove it yourself) that

left-parenthesis upper A union upper B right-parenthesis Superscript c Baseline equals upper A Superscript c Baseline upper B Superscript c Baseline and left-parenthesis upper A upper B right-parenthesis Superscript c Baseline equals upper A Superscript c Baseline union upper B Superscript c Baseline period

      Complements turn unions into intersections, and vice versa. These set-theoretic results are known as DeMorgan's laws. The results extend to infinite sequences. Given events upper A 1 comma upper A 2 comma ellipsis,

left-parenthesis union Underscript i equals 1 Overscript infinity Endscripts upper A Subscript i Baseline right-parenthesis Superscript c Baseline equals intersection Underscript i equals 1 Overscript infinity Endscripts upper A Subscript i Superscript c Baseline and left-parenthesis intersection Underscript i equals 1 Overscript infinity Endscripts upper A Subscript i Baseline right-parenthesis Superscript c Baseline equals union Underscript i equals 1 Overscript infinity Endscripts upper A Subscript i Superscript c Baseline period

       Example 1.29 Four dice are rolled. Find the probability of getting at least one 6.The sample space is the set of all outcomes of four dice rollsBy the multiplication principle, there are elements. If the dice are fair, each of these outcomes is equally likely. It is not obvious, without some new tools, how to count the number of outcomes that have at least one 6.Let be the event of getting at least one 6. Then the complement is the event of getting no sixes in four rolls. An outcome has no sixes if the dice rolls a 1, 2, 3, 4, or 5 on every roll. By the multiplication principle, there are possibilities. Thus, and

upper P left-parenthesis upper A union upper B union upper C right-parenthesis equals upper P left-parenthesis upper A right-parenthesis plus upper P left-parenthesis upper B right-parenthesis plus upper P left-parenthesis upper C right-parenthesis minus upper P left-parenthesis upper A upper B right-parenthesis minus upper P left-parenthesis upper A upper C right-parenthesis minus upper P left-parenthesis upper B upper C right-parenthesis plus upper P left-parenthesis upper A upper B upper C right-parenthesis period

      As we first include the sets, then exclude the pairwise intersections, then include the triple intersection, this is called the inclusion–exclusion principle. The proof is intuitive with the help of a Venn diagram, which we leave to the reader. Write

upper A union upper B union upper C equals left-bracket upper A union upper B right-bracket union left-bracket upper C left-parenthesis upper A upper C union upper B upper C right-parenthesis Superscript c Baseline right-bracket period

      The bracketed sets upper A union upper B and upper C left-parenthesis upper A upper C union upper B upper C right-parenthesis Superscript c are disjoint. Thus,

      Write upper C as the disjoint union

upper C equals left-bracket upper C left-parenthesis upper A upper C union upper B upper C right-parenthesis right-bracket union left-bracket upper C left-parenthesis upper A upper C union upper B upper C right-parenthesis Superscript c Baseline right-bracket equals left-bracket upper A upper C union upper B upper C right-bracket union left-bracket upper C left-parenthesis upper A upper C union upper B upper C right-parenthesis Superscript c Baseline right-bracket period

      Rearranging gives

upper P left-parenthesis upper C left-parenthesis upper A upper C union upper B upper C right-parenthesis Superscript c Baseline right-parenthesis equals upper P left-parenthesis upper C right-parenthesis minus upper P left-parenthesis upper A upper C union upper B upper C right-parenthesis period StartLayout 1st Row 1st Column upper P left-parenthesis upper A union upper B union upper C right-parenthesis 2nd Column equals upper P left-parenthesis upper A right-parenthesis plus <p style= Скачать книгу