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Modern Engineering Thermodynamics

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744 CHAPTER 18: Introduction to Statistical <strong>Thermodynamics</strong><br />

Probabilities of related events that are not mutually exclusive have the following interpretation. Assume that<br />

event B depends in some way on another event A. Then, the conditional probability of event B occurring given<br />

thefactthateventA has already occurred is written P B/A .Thecompound probability that both events A and B<br />

will occur is written as P AB . The relation between these two probabilities is the basic and logic probability statement,<br />

written as<br />

PAand ð BÞ = P AB = P A P B/A = P B P A/B (18.32)<br />

If events A and B are totally independent, their probabilities are uncoupled and P A/B = P A and P B/A = P B : Then,<br />

Similarly, the basic or logic probability statement can be written as<br />

PAand ð BÞ = P AB = P A P B (18.33)<br />

PAor ð BÞ = P A + P B − P AB (18.34)<br />

If events A and B are mutually exclusive (i.e., they cannot occur simultaneously), then P AB = 0andP (A or B) = P A + P B .<br />

EXAMPLE 18.6<br />

In the draw of a single card from a full deck of playing cards, what is the probability that it will be an ace or a spade?<br />

Solution<br />

Equation (18.34) gives P (ace or spade) = P ace + P spade − P ace of spades . Now,<br />

so<br />

P ace = 4<br />

52 = 1 13 , P spade = 13<br />

52 = 1 4 , P ace of spades = 1<br />

52<br />

P ðace or spadeÞ = 4<br />

52 + 13<br />

52 − 1<br />

52 = 16 = 0:308 = 30:8%<br />

52<br />

Exercises<br />

16. Instead of drawing either an ace or a spade in Example 18.6, determine the probability of drawing either a 2 or a 3 of<br />

any suit. Answer: P(2 or 3) = 0.154 = 15.4% of the time.<br />

17. What is the probability of drawing an ace of spades in Example 18.6? Answer: P(ace and spade) = 0.0192 = 1.92% of<br />

the time.<br />

18. Find the probability of drawing four aces in a row out of a standard 52 card deck. Hint: Initially, there are four aces<br />

scattered among the 52 cards, but after the first ace is drawn, only three aces remain among 51 cards, and so forth.<br />

Answer: P(ace and ace and ace and ace) = 3.7 × 10 −6 .<br />

Table 18.6 lists the probability of the different five-card poker hands.<br />

Table 18.6 Probabilities of Five-Card Poker Hands<br />

Name of Hand<br />

Number of Card Combinations<br />

Giving This Hand<br />

Probability of the Hand<br />

Royal flush 4 1.54 × 10 −6<br />

Straight flush 36 1.38 × 10 −5<br />

Four of a kind 624 0.00024<br />

Full house 3,744 0.00144<br />

Flush 5,108 0.00197<br />

Straight 10,200 0.00392<br />

Three of a kind 54,912 0.02113<br />

Two pair 123,552 0.0475<br />

One pair 1,098,240 0.42257<br />

All other hands 1,302,540 0.50118<br />

Total 2,598,960 1.00000

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