2.5 Sample Spaces Having Equally Likely Outcomes


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1 Sample Spaces Having Equally Likely Outcomes 3 Sample Spaces Having Equally Likely Outcomes Recall that we had a simple example (fair dice) before on equallylikely sample spaces Since they will appear often, we will discuss them more in detail in this section Consider an experiment with N possible outcomes Let s denote its sample space by If all outcomes are equallylikely then S = {s 1,s,,s N } P({s 1 })=P({s })= = P({s N })= 1 N since P(S)=1 So if E is an event with M elements in S, such that E = {s i1,s i,,s im } then This means that P({s i1 })+P({s i })+ + P({s im })= 1 N + 1 N N = M N in an equallylike sample space S the number of elements in E the number of elements in S Example 9 A fair dice is rolled P(an even number shows)= 3 6 Notation 1 To denote the number of elements in an event E, we ll write E Example 10 A pair of dice is rolled What is the probability that the sum of two shows 7? The event is that the sum of both dice equals 7 So it is and Also the size of the sample space is E = {(1,6),(,),(3,4),(4,3),(,),(6,1)} E = = 36 E 6 36 Example 11 3 balls are randomly drawn from an urn containing 6 white and black balls What is the probability of drawing one white and two black balls? Since there are 11 balls in total, The event of interest is S = {all possible 3ball selections} 11 = 16 3 E = {all possible 3ball selections where 1 is white and two are black} So
2 4 Axioms of Probability Choose 1 white out of Choose black out of = 6 4 = 60 E Example 1 A committee of is to be selected from a group of 6 men and 9 women If the selection is made randomly, what is the probability that the committee consists of 3 men and women? hence The event is and S = {all possible people groups}, 1 E = {all possible people groups consisting of 3 men and women} E = Choose 3 men out of Choose women out of Example A poker hand consists of cards If the cards have distinct consecutive values (1,,,10,J,Q,K) and are not all of the same suit, we say that the hand is straight What is the probability that one is dealt a straight? For example, is a straight, J Q is a straight, is not a straight S = {all card hands}
3 Sample Spaces Having Equally Likely Outcomes and so Our event is E = {straighthands} We need to count how many straight hands can be formed For this purpose, let s consider the hands consisting of only 1,,3,4, s How many straights are there in this case? Note that there are 4 of spades, 4 of hearts, 4 of diamonds and 4 of clubs Without restriction we have = 4 of 1?? 3? 4?? But among these we have 4 hands which are not straight,namely: is not a straight, is not a straight, is not a straight, is not a straight So we have 4 4 straights if we consider only 1,,3,4, cards Similarly, we have the same amount of straights for,3,4,,6 etc 1?? 3? 4??! (4 4) straights? 3? 4?? 6?! (4 4) straights 3? 4?? 6? 7?! (4 4) straights 4?? 6? 7? 8?! (4 4) straights? 6? 7? 8? 9?! (4 4) straights 6? 7? 8? 9? 10?! (4 4) straights 7? 8? 9? 10? J?! (4 4) straights 8? 9? 10? J? Q?! (4 4) straights 9? 10? J? Q? K?! (4 4) straights 10? J? Q? K? 1?! (4 4) straights
4 6 Axioms of Probability In total, there are 10(4 4) straight hands So 10(4 4) Example 14 In the game of bridge, the entire deck of cards is dealt out to 4 players What is the probability that 1 one of the players receives all of spades, each player receives one ace? 1 Consider only the "Player 1" The sample space: which leads to The event is S = {all card selections for "Player 1"} E 1 = {all card selections for "Player 1" where all are spades} Note that this possible only one way So E 1 = 1 P(E 1 )= 1 Since there are 4 players in the game, each have this probability to receive all spades our probability is P(one of the players receives all of spades)= 4 Now put the aces aside Then you need to divide 48 cards equally to 4 players So there are 48 ways Next distribute aces to players in 4! ways So the size of the event is 48 4! just different ways of distributing cards to 4 players equally, which can be done in,,, P(each player receives one ace)= 4! 48,,,
5 Sample Spaces Having Equally Likely Outcomes 7 Example 1 There are n people at a party What is the probability that no two people have the same birthday? (That is, no two of them were born on the same day of a year) the all possible birthdays for n people Then its size is {z 36 } nmany =36 n And the event is that no two shares the same day E = (36 {z n + 1) } nmany = 36! (36 n)! 36! (36 n)! 36 n We can ask the question, what is the probability that at least people at the party have the same birthday? Clearly it equals ! (36 n)! 36 n Interestingly, if you choose n = 0 this probability is approximately 097! Surprising?
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Math 166 Spring 2007 c Heather Ramsey Page 1 Math 166  Exam 2 Review NOTE: For reviews of the other sections on Exam 2, refer to the first page of WIR #4 and #5. Section 7.1  Experiments, Sample Spaces,
More information