Robert Bartoszynski - Probability and Statistical Inference

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Updated classic statistics text, with new problems and examples
Probability and Statistical Inference, Third Edition
Probability and Statistical Inference 

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Example 3.11 Matching Problem

A secretary typed картинка 1224letters and addressed картинка 1225envelopes. For some reason, the letters were put into envelopes at random. What is the probability of at least one match, that is, of at least one letter being put into the correct envelope?

Solution

This problem appears in many textbooks under various formulations (e.g., of guests receiving their hats at random). One could expect the probability of at least one match to vary greatly with картинка 1226. However, the contrary is true: this probability is almost independent of картинка 1227. Let картинка 1228be the event that th letter is placed in the correct envelope Using formula 26 we have By - фото 1229th letter is placed in the correct envelope. Using formula ( 2.6), we have

By symmetry the probability of each intersection depends only on the number of - фото 1230

By symmetry, the probability of each intersection depends only on the number of events in the intersection, 2, so we let denote the probability of the intersection of events Clearly the numbers of - фото 1231denote the probability of the intersection of events Clearly the numbers of terms in the consecutive sums are - фото 1232events, Clearly the numbers of terms in the consecutive sums are and 325 - фото 1233Clearly, the numbers of terms in the consecutive sums are

and 325 To evaluate we can argue as follows Assume - фото 1234

and

(3.25) To evaluate we can argue as follows Assume that the envelopes are ordered in - фото 1235

To evaluate картинка 1236we can argue as follows: Assume that the envelopes are ordered in some way. The total number of ways one can order картинка 1237letters is Probability and Statistical Inference - изображение 1238. If specific Probability and Statistical Inference - изображение 1239events, say Probability and Statistical Inference - изображение 1240are to occur (perhaps in conjunction with other events), then the letters number картинка 1241must be at their appropriate places in the ordering (to match their envelopes). The remaining Probability and Statistical Inference - изображение 1242letters can appear in any of the Probability and Statistical Inference - изображение 1243orders. Thus,

Probability and Statistical Inference - изображение 1244

Consequently, the th term in the sum 325 equals up to the sign and we obtain - фото 1245th term in the sum ( 3.25) equals (up to the sign)

and we obtain Since we have - фото 1246

and we obtain

Since we have with the accuracy increasing as - фото 1247

Since

we have with the accuracy increasing as The approximation is - фото 1248

we have

with the accuracy increasing as The approximation is actually quite good for - фото 1249

with the accuracy increasing as картинка 1250. The approximation is actually quite good for small картинка 1251. The limiting value is 0.63212056, while the exact values of the probability of at least one match for selected values of are E - фото 1252of at least one match for selected values of are Example 312 Ballot Problem Suppose that in an election candidate A - фото 1253are

Example 312 Ballot Problem Suppose that in an election candidate A receives - фото 1254

Example 3.12 Ballot Problem

Suppose that in an election, candidate A receives картинка 1255votes, while candidate B receives картинка 1256votes, where картинка 1257. Assuming that votes are counted in random order, what is the probability that during the whole process of counting A will be ahead of B?

Solution

Note that other votes, if any, do not matter, and we may assume that картинка 1258is the total number of votes. The process of counting votes is determined by the arrangement of the votes, that is, the arrangement of картинка 1259symbols A, and картинка 1260symbols B. Clearly, such an arrangement is uniquely determined by specifying the locations of the As (or, equivalently, Bs). It might be helpful to use a graphical representation: define the function картинка 1261as the net count for candidate A after inspection of картинка 1262votes. Thus, if in the first картинка 1263votes, we had votes for A and votes for B then We can then represent the process of count - фото 1264votes for A and votes for B then We can then represent the process of counting as a - фото 1265votes for B, then We can then represent the process of counting as a polygonal line that starts - фото 1266. We can then represent the process of counting as a polygonal line that starts at the origin and has vertices see Figure 32 Figure 32Process of counting votes In Figure 32 we have - фото 1267(see Figure 3.2).

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