# 1 Counting Problems From Fred Greenleaf’s QR book Compiled by Samuel Marateck.

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1 Counting Problems From Fred Greenleaf’s QR book Compiled by Samuel Marateck

2 P(success) = # of successful outcomes # of possible outcomes

3 Given an urn with 20 marbles 8 of which are green, what is the probability of choosing a green one?

4 P(green) = # of green marbles # of marbles P(green) = 8/20 or 40%

5 This also means that if you perform the experiment many times, let’s say 100 times, you’d expect to choose a green marble 40 times.

6 What is the probability of choosing a non- green marble?

7 What is the probability of choosing a non- green marble? P(non-green) = # of non-green marbles total # of marbles = 12/20 or 60%

8 Note the total probability sums to 100% P(non-green) + P(green) = 40%+60%=100%

9 Throw a pair of dice. What combinations can we get? 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 These are independent events in that the result of one throw does not effect the result of the next throw.

10 Which ones add to 7 or 11? 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 There are six that add to 7 and two that add to 11. There are 36 total outcomes.

11 What’s the probability of getting a 7 or 11? P(7 or 11) = # of 7 or 11 total # of outcomes In probability when used for independent Events, the or means use a “+”, so we add the probabilities. P(7 or 11) = P(7) + P(11) = 6/36 + 2/36 =8/36

12 A physicist says she will give you \$3 if you get a 7 or 11; but you will have to give her \$1 for any other outcome. Are these fair odds?

13 Let’s say you throw 36 times. You’d expect to win 8x3 or 24 dollars but lose 36 – 8 or 28 times and have to pay \$28. So you’d loose \$4 in 36 throws.

14 What are fair odds? Let w be the amount you get if you win and v be the amount you must pay if you lose. So 8w = 28v or w/v = 28/8 = 3.5. If you give her \$1 if you lose, she must give you \$3.50 if you win

15 How many two letter words can be formed using the Roman alphabet, where any two letters can form a word?

16 Let’s say you choose an A first. So you can choose 26 letters as your second letter. But you can choose 26 letters as your first one. And for each of these you can chose 26 for your second one. So the total is 26x26.

17 This is an example of a list with replacement since you can use a letter again once it has been used. It is also an ordered list since the order is important. The word AB is different from the word BA.

18 For a five letter word, you can choose 26 letters for each spot, so the result Is 26 5.

19 How many 5 letter words can you compose If no letters are repeated? This is an example of an ordered list without replacement, since once a letter has been used it cannot be used again.

20 For your first choice you have 26 letters. For each of these letters you have 25 letters that have not been chosen. And for each of these you have 24 that have not been chosen and so on. So the answer Is 26x25x24x23x22.

21 20 players are in a tournament. The top 5 players will be ranked, first, second, etc. How many ways can they be ranked? This is an example of an ordered list with without replacement.

22 If player A is ranked first, then there are 19 players who can be ranked second. And for each of these there are 18 who can be ranked third and so forth.

23 So the answer is 20x19x18x17x16.

24 You have a deck of 13 cards and you draw a card from the deck. You then replace it. What is the probability of drawing a different card from the deck?

25 You have a deck of 13 cards and you draw a card from the deck. You then replace it. What is the probability of drawing a different card from the deck? Since there are 12 cards that have not been drawn, the probability is 12/13 or 92%.

26 If three cards are drawn with replacement from this deck, what is the probability they will all be different?

27 If three cards are drawn with replacement from this deck, what is the probability they will all be different? 12/13 * 11/13 or 78%

28 If four cards are drawn with replacement from this deck, what is the probability they will all be different?

29 If four cards are drawn with replacement from this deck, what is the probability they will all be different? 12/13 * 11/13 * 10/13 or 60%

30 If five cards are drawn with replacement from this deck, what is the probability they will all be different?

31 If five cards are drawn with replacement from this deck, what is the probability they will all be different? 12/13 * 11/13 * 10/13 * 9/13 or 42%

32 If six cards are drawn with replacement from this deck, what is the probability they will all be different?

33 If six cards are drawn with replacement from this deck, what is the probability they will all be different? 12/13 * 11/13 * 10/13 * 9/13 * 8/13 or 26%

34 You have a deck of 5 cards and you draw 5 cards from the deck replacing them each time. What is the probability of drawing 5 different card from the deck?

35 1*4/5*3/5*2/5*1/5 =.0384

36 P(drawing 2 or more that are the same)?

37 1 -.0384 = 96%

38 Combination Symbols ( n m ) this is the symbol for n take m. How many committees of 3 can be chosen from 5 ? This is an example of a unordered list without replacement. The order that people appear on a committee is not important; but once a person is chosen he cannot be chosen again.

39 This is ( 5 3 ) or 5*4*3/(3*2*1) or 20/2 or 10. Can we write this using factorials?

40 If you write this with factorials, you have to complete 5*4*3 with 2*1; but this means you have to divide by 2*1. So we get 5*4*3*2*1/(3*2*1 * 2*1) or 5!/(3! * 2!).

41 What is ( n m ) using factorials?

42 In general ( n m ) is written as n!/(m! * (n-m)!)

43 Now ( 5 3 ) = ( 5 2 ), what is the equivalent of ( n m )?

44 The equivalent of ( n m ) is ( n n - m ).

45 If you flip a coin 5 times, how many different permutations can you get?

46 If you flip a coin 5 times, how many different permutations can you get? There are two choices for each coin, H or T. So the number of permutations are 2 5, or 32. How would you categorize this?

47 It is an ordered list with replacement. For instance, HTHTH is different from THTHT. It is with replacement since once a coin lands as a head, it can land as a head again.

48 How many times would you expect to see 5 heads?

49 How many times would you expect to see 5 heads? ( 5 5 ) since you are taking 5 things from 5

50 How many times would you expect to see 4 heads?

51 How many times would you expect to see 4 heads? ( 5 4 )

52 How many times would you expect to see 3 heads?

53 How many times would you expect to see 3 heads? ( 5 3 )

54 How many times would you expect to see 2 heads?

55 How many times would you expect to see 2 heads? ( 5 2 )

56 How many times would you expect to see 1 head?

57 How many times would you expect to see 1 head? ( 5 1 )

58 How many times would you expect to see 0 heads?

59 How many times would you expect to see 0 heads? ( 5 0 )

60 What do these add up to? ( 5 0 ) + ( 5 1 ) + ( 5 2 ) + ( 5 3 ) + ( 5 4 ) + ( 5 5 ) 1 + 5 + 10 + 10 + 5 + 1 = 32

61 The probability of each throw is ( 5 0 )/32 + ( 5 1 )/32 + ( 5 2 )/32 + ( 5 3 )/32 + ( 5 4 )/32 + ( 5 5 )/32 1/32 +5/32+10/32+10/32+5/32+1/32 =32/32 Or 100%

62 If you choose two cards from a deck, what is the probability that they are both aces? How do you categorize this?

63 This is an unordered list without replacement, since once a card has been drawn, it cannot be drawn again.

64 If you choose two cards from a deck, what is the probability that they are both aces? P(1 ace) = 4/52 P(2 aces)?

65 Since there are 3 aces left in the deck of 51 cards, P(2nd ace) = 3/51 and the P(2 aces) = 4/52*3/51

66 If you choose two cards from a deck, what is the probability that they both have the same face value?

67 We saw that P(2 aces) = 4/52*3/51, so P(2 kings ) = 4/52*3/51 P(2 queens ) = 4/52*3/51 for all 13 categories. P(2 of the same category)= 13* 4/52*3/51

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