Probability and Statistical Inference. Robert Bartoszynski

Probability and Statistical Inference

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the total number of lines from images

minus the number of lines from images

which touch or cross the images

‐axis. The total number of lines leading from images

, since each such line has images

steps “up” and images

steps “down,” which can be ordered in any manner. Thus, it remains to count the number of lines from images

that touch or cross the images

‐axis. Let images

be the set of all such lines. Each line in images

must touch the images

‐axis for the first time at some point, say images

(see Figure 3.3). If we reflect the part of this line that lies to the left of images

with respect to images

‐axis, we obtain a line leading from images

. Moreover, different lines in images

will correspond to different lines leading from images

and each line in the latter set will be obtained from some line in images

. This means that the set images

has the same number of lines as the set of lines leading from images

. But the latter set contains images

lines, since each such line must have images

steps “up” and images

steps “down.” Consequently, the required probability equals

Example 3.13 Poker

We now consider the probabilities of several poker hands (some students will probably say that finally the book gives some useful information).

In poker, five cards are dealt to a player from a standard deck of 52 cards. The number of possible hands is therefore images . The lowest type of hand is that containing one pair (two cards of the same denomination, plus three unmatched cards). To find the number of possible hands containing one pair, one can think in terms of consecutive choices leading to such a hand:

1 The denomination of the cards in a pair can be chosen in ways.

2 The suits of the pair can be chosen in ways.

3 The choice of denominations of the remaining three cards can be made in ways.

4 The suits of those three cards may be chosen in ways. Altogether, combining (a)–(d), we have

The next kind of hand is the one containing two pairs. Here the argument is as follows:

1 The denominations of the two pairs can be selected in ways.

2 The suits of cards in these two pairs can be selected in ways.

3 The remaining card may be chosen in ways (two denominations are eliminated).

Combining (a)–(c), we have

Finally, we calculate the probability of a straight (probabilities of remaining hands are left as exercise). A straight is defined as a hand containing five cards in consecutive denominations but not of the same suit (e.g., 9, 10, jack, queen, and king). An ace can appear at either end, so we could have a straight of the form ace, 2, 3, 4, 5, as well as 10, jack, queen, king, ace.

The number of hands with a straight can be computed as follows: Each such hand is uniquely determined by the lowest denomination (ace, images ) in 10 ways. Then, the suits of five cards are chosen in images ways: images is the total number of choices of suits, and we subtract 4 selections in which all cards are of the same suit. Thus,

Problems

1 3.3.1 (i) A committee of size 50 is to be formed out of the US Senate at random. Find the probability that each state will be represented. (ii) If a committee of size is to be formed out of the US Senate find how large must be in order for the event “at least one senator from Ohio is included” to be more likely than the event “no senator from Ohio is included.”

2 3.3.2 A shipment of 30 items is received. For the quality control, three items are randomly selected, and if more than one of them is defective then the whole shipment is rejected. Find the probability that the shipment will be accepted if it has: (i) Three defective items (ii) Ten defective items.

3 3.3.3 How many ways can one order the deck of 52 cards so that all four kings are next to each other?

4 3.3.4 Peter lives at the corner of 2nd Avenue and 72nd Street. His office is in the building at a corner of 7th Avenue and 78th Street. The streets and avenues in the city form a perpendicular grid, with no streets or passages in the middle of the blocks. Peter walks to work along either street or avenue, always in the direction that gets him closer to his office. He always returns home by subway, so he walks across town only once a day. (i) How many different paths can Peter choose to go to work? (ii) If Peter makes a list of all possible paths and chooses one of them randomly every morning, how likely it is that he will not walk 4th Avenue between 75th and 76th streets during the next five working days?

5 3.3.5 (Poker Hands) (Poker

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