Probability and Statistical Inference. Robert Bartoszynski

Probability and Statistical Inference

following hands: (i) Royal flush (ace, king, queen, jack, and 10 in one suit), (ii) Straight flush (five cards of one suit in a sequence, but not a royal flush), (iii) Flush (five cards in one suit, but not a straight flush nor a royal flush), (iv) Four‐of‐a‐kind (four cards of the same denomination), (v) Full house (one pair and one triple of the same denomination), (vi) Three‐of‐a‐kind (three cards of the same denomination plus two cards unmatched).

6 3.3.6 Find the probability that a poker hand will contain two pairs (one red and the other black) and one unmatched card.

7 3.3.7 A poker player has Q. He discards and Q and obtains 2 cards.3(i) What is the probability that he will have a straight? (ii) Answer the same question if Q is replaced by J (i.e., he discards and J).

8 3.3.8 A poker player has Q. She discards and Q and obtains 2 cards. What is the probability that she will have: (i) A straight flush. (ii) A flush, but not a straight flush. (iii) A straight, but not a straight flush.

9 3.3.9 A poker player has three‐of‐a‐kind. He discards the two unmatched cards and obtains two new cards. Find the probability that he will have: (i) Three‐of‐a‐kind. (ii) Four‐of‐a‐kind. (iii) A full house.

10 3.3.10 (i) If balls are put at random into boxes, find the probability of exactly one box remaining empty? (ii) If balls are randomly placed into boxes (), labeled , find the probability that no box is empty.

11 3.3.11 Compute probabilities of winning numbers in lotteries, where the player chooses: (i) 5 out of 44 numbers. (ii) 6 out of 55 numbers.

12 3.3.12 Find the number of polygonal lines with vertices , where is as in Example 3.12 and with possible edges leading from to or , connecting the points: (i) and . (ii) and . iii) and .

13 3.3.13 Find the number of polygonal lines (as in Problem 3.3.12) that join the points (2,3) and (16, 5) and: (i) Never touch the ‐axis. (ii) Never touch the line = 7.

3.4 Multinomial Coefficients

Choosing a subset of size images out of a set of size images is logically equivalent to partitioning the set of size images into two subsets, one of size images and the other of size images . The number of such partitions is, by definition,

The theorem below generalizes this scheme.

Theorem 3.4.1 Let images be positive integers such that images . The number of ways a set of images elements can be partitioned into images subsets of sizes images equals

(3.26) equation

Proof: A partition above can be accomplished in steps: First, we choose images out of images elements to form the first subset of the partition. Next, we choose images elements out of the remaining images elements, and so on, until we have images elements, from which we choose images to form the next‐to‐last subset. The remaining images elements form the last subset This can be accomplished, in view of Theorem 3.2.2, in

(3.27) equation

ways. Simple algebra shows that formula (3.27) is the same as formula (3.26).

The ratio (3.26) is called multinomial coefficient and is denoted by

As a generalization of Newton's binomial formula, we have

Theorem 3.4.2 For every integer images ,

(3.28) equation

where the summation is extended over all subsets images of nonnegative integers with images .

Proof: In the product images , one term is taken from each factor so that the general term of the sum has the form images with images . From Theorem 3.4.1, it follows that the number of times the product images appears equals (3.26).

In an analogy with a formula (3.17), the sum of all multinomial coefficients equals images , which follows by substituting images in (3.28).

The theorem is illustrated by the following example:

Example 3.14

Suppose that one needs the value of the coefficient of Скачать книгу