a The 12 companies are distinguishable repetitions are not possible and arrange

# A the 12 companies are distinguishable repetitions

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(a) The 12 companies are distinguishable, repetitions are not possible, and arrange- ments are irrelevant, so counting rule 3 is appropriate. The number of combinations of 12 companies taken four at a time is ( 12 4 ) = 495. (b) The number of ways of selecting four gold-mining companies is ( 5 4 ) = 5, and therefore Pr(4 gold-mining companies) = 5 / 495. The number of ways of selecting four industrial companies is ( 7 4 ) = 35, with probability 35 / 495. The probability of an undiversified portfolio is the sum of these two probabilities: (5 + 35) / 495 = 0 . 081, about “one chance in 12”. Example 30B: At a party, there are substantial stocks of five brands of beer — Castle, Lion, Ohlssons, Black Label and Amstel. One of the party-goers grabs two cans without looking. How many different combinations of 2 beers are possible? The brands are distinguishable, repetitions are permitted, but the ordering is unim- portant, so counting rule 5 is used. The number of ways of selecting two cans from five brands allowing repetitions is parenleftbigg 5 + 2 1 2 parenrightbigg = parenleftbigg 6 2 parenrightbigg = 15 . Note, that not all of these 15 outcomes are equally probable! Example 31B: A group of 20 people is to travel in three light aircraft seating 4, 6 and 10 people respectively. What is the probability that three friends travel on the same plane? The total number of ways of choosing combinations of sizes 4, 6 and 10 from a group of size 20 is, by counting rule 7, given by parenleftbigg 20 4 6 10 parenrightbigg = 20! 4! × 6! × 10! = 38 798 760 . If the three friends travel in the four-seater aircraft, the remaining 17 must be split into groups of sizes 1, 6 and 10. This can be done in ( 17 1 6 10 ) ways. Similarly, if they travel in the six-seater, the other 17 must be split into groups of 4, 3 and 10,and if they travel in the 10-seater, the others must be in groups of sizes 4, 6 and 7. Thus the total number of ways the three friends can be together is parenleftbigg 17 1 6 10 parenrightbigg + parenleftbigg 17 4 3 10 parenrightbigg + parenleftbigg 17 4 6 7 parenrightbigg = 4 900 896 ways . Thus, Pr(3 friends together) = (4 900 896) / (38 798 760) = 0 . 1263. Example 32B: A bridge hand consists of 13 cards dealt from a pack of 52 playing cards. What is the probability of being dealt a hand containing exactly 5 spades? The cards are distinguishable, repetitions are not possible, and the arrangement of the cards is irrelevant (the order in which you are actually dealt the cards does not make any difference to the hand). So counting rule 3 is the one to use to determine that the total number of possible hands is ( 52 13 ) . Applying counting rule 3 twice more, the number of ways of drawing 5 spades from the 13 in the pack and the remaining 8 cards for the hand from the 39 clubs, hearts and diamonds is ( 13 5 )( 39 8 ) . Hence
CHAPTER 3. PROBABILITY THEORY 73 Pr(5 spades) = parenleftbigg 13 5 parenrightbiggparenleftbigg 39 8 parenrightbigg parenleftbigg 52 13 parenrightbigg = 0 . 1247 .

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