Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Combination Symbols A supplement to Greenleafs QR Text Compiled by Samuel Marateck ©2009.

Similar presentations


Presentation on theme: "1 Combination Symbols A supplement to Greenleafs QR Text Compiled by Samuel Marateck ©2009."— Presentation transcript:

1 1 Combination Symbols A supplement to Greenleafs QR Text Compiled by Samuel Marateck ©2009

2 2 How many 4-card hands consisting of 1 king and 3 queens can be chosen from a deck?

3 3 How many 4-card hands consisting of 1 king and 3 queens can be chosen from a deck? Since order does not matter and there are four kings and four queens in the deck, the answer is: ( 4 1 ) ( 4 3 )

4 4 What is the meaning of ( 4 1 )? Its the number of ways we can choose one thing from four, independent of the order. It is pronounced four choose one.

5 5 Similarly ( 4 3 ) is the number of ways we can choose three things from four independent of the order. It is pronounced four choose three.

6 6 In ( 4 1 ) ( 4 3 ), why do we multiply the two?

7 7 For each king there are three queen pairings. These are the pairings for the king of spades: k Q Q Q

8 8 But there are also k, k and k. So there are 16 different combinations, four for each King.

9 9 What is the probability of choosing 4-card hands consisting of 1 king and 3 queens from a deck?

10 10 What is the probability of choosing 4-card hands consisting of 1 king and 3 queens from a deck? ( 4 1 ) ( 4 3 ) / ( 52 4 )

11 11 We divide by ( 52 4 ) since this is the number of ways we can choose four cards at random from a deck.

12 12 Lets evaluate ( 4 1 ) ( 4 3 ) / ( 52 4 )

13 13 ( 4 1 ) ( 4 3 ) / ( 52 4 ) is: 16/(52*51*50*49/(4*3*2*1)) = or.006%

14 14 Out of how many hands would you expect to get this hand?

15 15 Out of how many hands would you expect to get this hand? is 6x 10 -5, so in 10 5 hands you would expect to get 6 such hands or in one out of 16,666 hands you would get this hand.

16 16 How many 5-card hands can you get that have three aces?

17 17 How many 5-card hands can you get that have three aces? The number of ways we can choose three aces is ( 4 3 ). How many cards are left in the deck?

18 18 How many non-aces are in the deck? There are 48 non-aces left in the deck and there are two more cards to choose for our hand.

19 19 So there are ( 4 3 ) ( 48 2 ) ways we can get three aces: 4*48*47/2 = 4*47*24 = 4512 ways.

20 20 What is the probability of getting three aces in a 5-card hand?

21 21 What is the probability of getting three aces in a 5-card hand? ( 4 3 ) ( 48 2 ) / ( 52 5 ) = 4512/((52*51*50*49*48)/(5*4*3*2*1)) = 4512/ =.00174

22 22 What is the probability of winning the lottery?

23 23 What is the probability of winning the lottery? There are 54 numbers that you can choose from; the numbers 1 to 54. You must choose the five correct numbers independent of their order. The answer is:

24 24 P(winning) = 1/( 54 5 ) ( 54 5 ) = 54*53*52*51*50/120 1/( 54 5 ) = 3.16 x 10 -7

25 25 If there are 6 pegs distributed in a circle and a line is drawn from each peg to each other peg, how many lines are there?

26 26 For each peg 5 lines are drawn; but there are 6 pegs. Since, however, each line connects two pegs, we are overcounting by 2, so we must divide by 2. What is the answer?

27 27 # of lines is 5*6/2 or 15.

28 28 Another way of looking at this is: From the first peg, 5 lines are drawn. From the second peg, 4 lines are drawn since it is already connected to the first peg. From the third peg, 3 lines are drawn, since it is connected to the first two, and so on,

29 29 For the six pegs, or 15 lines are drawn. For n pegs n-1 + n-2 + n lines are drawn. We know what the sum from 1 to m is.

30 30 The sum is: m(m+1)/2. Substituting n-1 for m, the sum from 1 to n-1 is (n-1)(n-1 +1)/2 =?

31 31 (n-1)(n-1 +1)/2 = n(n-1)/2 which is the answer we got before. Can we do this with combination symbols?

32 32 If there are 6 pegs distributed in a circle and a line is drawn from each peg to each other peg, how many lines are there?

33 33 There are 6 slots: How many ways can we place two item in these slots?

34 34 How many ways can we place two item in these slots? The answer is ( 6 2 ). For n pegs its ?

35 35 For n pegs its ( n 2 ).

36 36 How many ways can we choose a 5-card hand so that no two cards have the same face values?

37 37 How many ways can we choose a 5-card hand so that no two cards have the same face values? For the first card we have ( 52 1 ) ways we can choose the first card. How many choices do we have for the second card?

38 38 How many choices do we have for the second card? 48, since one face value has been eliminated. So the number of ways we can choose the second card is:

39 39 ( 48 1 ). The third card is?

40 40 ( 44 1 ). So the final answer is: ( 52 1 ) ( 48 1 )( 44 1 ) ( 40 1 ) ( 36 1 ). What is the probability?

41 41 P(each card has a different face value) = ( 52 1 ) ( 48 1 )( 44 1 ) ( 40 1 ) ( 36 1 ) ( 52 5 )

42 42 In a class of 25, what is the probability that two or more people have the same birthdate?

43 43 In a class of 25, what is the probability that two or more people have the same birthdate? We will first calculate the probability that no one has the same birthdate.

44 44 Given the first person, the probability that the second one has a different birth date is 364/365. That the first, second and third ones have different birth dates is: 1* 364/365*363/365. For all 25 people?

45 45 For all 25 people? P(different birth dates) = 364*363*362*361…341/ = 0.47

46 46 P(2 or more have same birth dates) =.53

47 47 There are 25 people to be chosen for a Committee or 5. What is my probability of being chosen?

48 48 What is the probability of my being chosen? ( 1 1 ) ( 24 4 )/ ( 25 5 ).

49 49 An urn contains 10 red balls and 40 black ones. What is the probability you will draw 2 red balls.

50 50 ( 10 2 ) ( 40 0 )/ ( 50 2 ) = 10*9/2 /(50*49/2) = 45/1225

51 51 An urn contains 17 red balls and 33 black ones. What is the probability you will draw 7 red balls if you choose 10 randomly?

52 52 ( 17 7 ) ( 33 3 )/ ( )

53 53 A jury pool contains 98 men and 75 women. 12 jurors are chosen at random. What is the probability that 6 will be women

54 54 ( 98 6 ) ( 75 6 )/ ( )


Download ppt "1 Combination Symbols A supplement to Greenleafs QR Text Compiled by Samuel Marateck ©2009."

Similar presentations


Ads by Google