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Given a set of outcomes S and a subset of them E (the “event”), then p(E) is the sum over sÎS of p(s). The old definition applied only when S was composed of equally likely outcomes, in which case, p(s) = 1/|S|. This new definition is more general.
Suppose a die is loaded so that p(6) =2/7 and p(n)=1/7 for n between 1 and 5. What is the probability of an even number being thrown on the die? It is p(2)+p(4)+p(6)= (1/7) + (1/7) + (2/7) = 4/7.
If ~E is the complement of E in S, (i.e., ~E = S-E), then p(~E) = 1-p(E). If E1 and E2 are two events of S (i.e., E1 Í S and E2 Í S), then p(E1 È E2) = p(E1) + p(E2) – p(E1ÇE2).
If a coin is loaded so that heads comes up 55 times out of a hundred throws, what is the probability that it will come up tails? It is p(tails) = 1-p(heads) = 45/100.
What is the probability that a loaded coin comes out heads if heads is three times more likely than tails? p(Heads)=3*p(Tails) and p(Heads)+p(Tails)=1, so p(Heads)=3/4.
What is the probability that a loaded die comes out 5 if rolling either 1 or 6 is three times more likely than rolling a 2 or rolling a 3, etc. that is p({1,6}) = 3*p(2) = 3*p(3) = 3*p(4)= 3*p(5). Thus, we divide the probability pie into 3+4= 7 slices, so p(5)=1/7.
A pair of loaded dice are rolled. P(4)=p(3)=2/8. The other face occur with probaility 1/8. What is the probability that the sum of the numbers on two dice is 7? We need to count the number of ways that 7 can be rolled and sum their probabilities. P([1,6])+p([2,5])+p([3,4])+p([4,3])+p(5,2])+p([6,1]) = [(1/8)*(1/8)] + [(1/8)*(1/8)] + [(2/8)*(2/8)] + [(2/8)*(2/8)] + [(1/8)*(1/8)] + [(1/8)*(1/8)] = 12/64.
Let E and F be two events in S with p(F)>0. If we know that F has occurred, then how does that affect the probability of E? Because F has occurred, the only way for E to occur is for an event in EÇF to occur. Moreover, we know that the old probabilities p(s) for sÎE are no longer accurate, because they assumed that any sÎS could occur, but now we know that only sÎF can occur. However, we can reuse the old values of p(s) by defining p(E|F) as p(EÇF)/p(F) where p(E|F) is pronounced “the probability of E given F.”
Suppose we have a set of men and women of varying heights. Let us use M67 to represent a man who is 67 inches tall. Our set of men and women is represented as {M67, W69, W60, M74, W49, M75}. What is the probability that a person chosen at random from this set has a height less than 6 feet (72 inches)? The subset is {M67, W69, W60, W49}, so the probability is 4/6=2/3. Now suppose we select a person at random and note that the person is male. Now what is the probability that the person is shorter than 6 feet? Only one of the three men is shorter than 72 inches, so we know intuitively that the probability is 1/3. Applying the definition, p( Short-man | man) = p(Short-man Ç man) / p(man) = p(M67) / p({M67, M74, M75}) = (1/6) / [(1/6)+(1/6)+(1/6)] = (1/6) / (1/2) = 2/6 = 1/3.
If E and F are two events in S, and knowing that F occurs does not change the probility that F occurs, then we say that E and F are independent. More formally, E and F are independent if and only if p(EÇF) = p(E)*p(F). This implies that p(E|F) = p(EÇF)/p(F) = p(E)*p(F)/p(F) = p(E).
If you roll two dice, and one falls off the table before you can see how it came out, then knowing how the one on the table came out tells you nothing about how the one on the floor came out. Thus, intuitively, if values on the two dice are independent. Thus, if E is the probability that the die on the table came out even and F is the probability that the die on the floor came out even, then p(E|F) = p(E). The probability that they both came out even is, intuitively ¼, and just as intuitition predicts, the definition of independence says p(EÇF)=p(E)*p(F) = ½* ½ .
Suppose that a pair of dice is loaded so that snake eyes comes up more frequently than usual. In particular, p(1) is three times more likely than p(n) for n between 2 and 6. What is the probability of snake eyes when two such dice are rolled? Since p(1) is three times more likely than the other 5 faces, each of the other faces has p(n)=1/(5+3)=1/8 and p(1)=3/8. The probability of rolling snake eyes (both dice come up 1) is p(1)*p(1)=(3/8)^2 = 9/64.
Suppose we define a “trial” as some outcome space S, and we define some event EÍS as a success. Moreover, the probabilities of E are the same every time we run a trial. Such trials are called Bernoulli trials, because the probability of success on one is independent of the probability of success on any other. If we run the same trial n times, the probability of getting k successes is C(n,k)p(E)kp(~E)n-k. We often want to know how this probability varies with k (e.g., we want to bet on k).
Let S be thowing a die, and let success be throwing a 1 or a 6. What is the probability of 10 successes in 12 trials? C(n,k)p(E)kp(~E)n-k = C(12,10)(1/3)10(2/3)2.
What is the probability that exactly 4 heads appear when a fair coin is flipped 5 times? C(5,4)(1/2)5(1/2)1.
What is the probability that exactly 3 heads appears when a fair coin is flipped 5 times, given that the first flip came up heads? This is just the probability that 2 heads appear when 4 are flipped, so it is C(4,2)(1/2)2(1/2)2.
What is the probability that exactly 4 head appear when a fair coin is fliipped 5 times, given that the first flip is came up tails? Just 4 heads in a row, so (1/2)4.
What is the probability that two people have the same birthday (assume there are 366 possible dates)? There are 366*366 events in S. Of these pairs, 366 represent two people having the same birthday, so the probability is 366/(366*366)= 1/366. Another way to show this is to recognize this as conditional probability of a person E have the same birthday as person F, given that F has a certain birthday. That is, the answer is p(E|F) = p(EÇF) / p(F) = (1/366*366) / (1/366) = 1/366.
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.