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Given a set S of equally likely outcomes, S, and EÍS, then p(E) = |E| / |S|. E is called an event and p(E) is the probability of the event E.
Practice:
What is the probability that a randomly selected integer chosen from the first 100 positive integers is odd? Let S = {1,2,3,…,100} and let E = {1,3,5,…,99}. Then p(E) = |E| / |S| = 50 / 100 = ½.
What is the probability that the sum of numbers on two dice is even when they are rolled? Let S = {1,2,3,4,5,6} ´ {1,2,34,5,6}. Then |S| = 36. Let E = {(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), …}. For each choice of 6 numbers on the first die, there are 3 numbers on the second die that add to an even number. Thus, half the 36 pairs add to even numbers. P(E) = |E| / |S|.
What is the probability that a coin lands heads-up 6 times in a row? Let S = {H,T}^6. Then |S| = 2^6 = 64. Let E = {HHHHHH}. Then |E| = 1. Thus, p(E) = |E|/|S| = 1/64.
What is the probability that a 5-card poker hand contains the ace, two, three, four and five of spades? S= all hands so |S|=C(52,5). E = one hand so |E| = 1. Thus, p(E) = 1/C(52,5).
What is the probability that a 5-card poker hand does not contain the queen of hearts? Let S = all possible poker hands = all 5-element subsets of a set of 52 objects. |S| = C(52,5). Let E = all poker hands that do not contain the queen of hearts = all 5-element subsets of a set of 51 objects. |E| = C(51,5). Thus, p(E) = |E|/|S| = C(51,5) / C(52,5) = (51!/(5!*46!)) / (52! / (5!*47!)) = (1 / (1*1)) / (52 / (1*47)) = 47/52.
What is the probability that a 5-card poker hand contins two pair? To build E, we can choose any two kinds (kind Î {ace, 2, 3, 4, 5, 6, 7, 8 , 9, 10, J, Q, K} for the pairs, and any C(4,2) for each pairs’ suits. Thus, there are C(13,2)*C(4,2)*C(4,2) possible pairs. After choosing the pairs, we cannot choose either kind for the remaining card, so there are only 52-8=44 choices left. Thus, there are C(44,1) choices for the remaining card. Thus, p(E) = C(13,2)*C(4,2)*C(4,2)*C(44,1) / C(52,5).
Given a set of outcomes S and an event EÍS, then the complement of E in S is S-E, and its probability is 1-p(E).
Given a set of outcomes S and two events E1ÍS and E2ÍS, then p(E1ÈE2) = p(E1) + p(E2) – p(E1ÇE2).
What is the probability that a 5-card poker hand contains at least 1 ace? Let E = hands with 1, 2, 3 or 4 aces, so ~E (the complement of E, which should be written with a bar over the E) = hands with 0 aces = hands drawn from deck without aces. Then |~E| = C(48,5). Thus, p(E) = 1-p(~E) = 1- C(48,5)/C(52,5).
What is the probability that a 5-card poker hand contains a flush, that is, 5 cards of the same suit? Let Ehearts = all cards are hearts. Then there are C(13,5) choices for Ehearts. Similarly, there are C(13,5) choices for Espades, Eclubs and Ediamonds. A flush is Eheart È Espades È Eclubs È Ediamonds, and the 4 sets of hands are disjoint, so p(flush) = 4*C(13,5)/C(52/5).
What is the probability that a 5-card poker hand contains a straight, that is, 5 consecutive kinds? Note that a straight can start or end with an ace. Thus, the lowest straight is (A, 1, 2, 3, 4) and the highest straight is (10, J, Q, K, A). Thus, there are 10 different types of straights. For each type, you get a choice of suits for each card. Thus, |(A, 1, 2, 3, 4)| = 4^5. We want the probability of any straight, so we can take the union of the events of each of the 10 straights. They are disjoint events, so we don’t have to subtract off their intersections. Thus, p(E) = 10*4^5 / C(52,5).
What is the probability that a 5-card poker hand contains a straight flush, that is, 5 consecutive kinds? As in calculating the straights, there are 10 types, but this time, each type has just 4 variants, one for each suit. Thus, p(E) = 10*4/C(52,5).
What is the probability that a 5-card poker hand contains cards of 5 different kinds? We can choose the kinds first, then for each 5-card combination of kinds, we can choose suits. Thus, there are C(13,5) types (ignoring suits) and 4 choices of suits for each card, so there are C(13,5)*4^5 hands in E. Thus, p(E)= C(13,5)*4^5 / C(52/5).
What is the probability that a 5-card poker hand contains cards of 5 different kinds and does not contain a straight or a flush? Our plan is to count the number of hands with distinct cards (given above) and subtract off the number of straights and the number of flushes. However, we be sure not to double count the straight flushes. Fortunately, all straights have distinct kinds and all flushes have distinct kinds, so they are subset of all hands with distinct kinds. Thus, E = |hands with distinct cards| - |straights or flushes with distinct cards| = 4*C(13,5) - |straights or flushes| = 4*C(13,5) - [ |straights| + |flushes| - |straight flushes| ] = 4*C(13,5) – [ 10*4^5 + 4*C(13,5) – 4*10].
What is the probability that a die never comes up an even number when rolled 6 times? S = {1,2,3,4,5,6}^6 and E = {1,3,5}^6, so p(E) = 3^6/6^6 = (3/6)^6 = (1/2)^6 = 1/64.
Find the probability of winning the lottery by selecting the correct six integers, where the order in which these integers are selected does not matter and the six integers are distinct, from the positive integers not exceeding 50. There is just one combination that is correct, so |E|=1. There are C(50,6) ways to choose 6 objects from 50, so the p(win) = 1/ C(50,6).
Find the probability of winning the lottery by selecting six of the correct 10 integers, where all numbers are distinct and unordered. The 10 numbers are choosen from 60. Any size-6 subset of the 10 is correct, so |E| = C(10,6). There are C(60,6) ways to choose the 6 numbers, so p(win) = C(10,6)/C(60,6).