Introduction to Probability

1.2.2 Interpretations If a bookmaer quotes payoff odds of 99 to 1 against a particular horse winning race, does that suggest the chance that the horse will win is 1/100, less than 1/100, or more than 1/100?

Odds to probability is a/a+b a = 1, b =99 1/1+99 = 1/100 but it is less than this because the house always wants the odds in their favor.

2.1.10 Suppose a fair coin is tossed n times. Find simple formulae in terms of n and k for a) P(k-1 heads | k-1 or k heads) b) P(k heads | k-1 or k heads)

just check the ******* answers

1.4.6 Suppose two cards are dealt from a deck of 52. What is the probability that the second card is a spade given that the first card is black?

Denote by B1 the event that first card is black and by S2 the event that the secondcard is a spade. We are looking for P(S2/B1) = P(B1S2)P(B1). Now, P(B1) = 1/2. At the same time, the event B1S2 can be partitioned as "spade-spade or club-spade". Therefore, P(B1S2) = (13 × 12 + 13 × 13)/ (52 × 51) = (25) /(4 × 51). This leads to: P(S2/B1) = (25) /(2 × 51) = 25/102.

Binomial Distribution P(k successes in n trials)

P(k successes in n trials) = (n k) p^kq^(n-k)

1.3.14 Show that P(A ∩ B) >= P(A) + P(B) - 1

Using the principle of inclusion-exclusion we have 1 ≥ P(A ∪ B) = P(A) + P(B) − P(A ∩ B). We isolate P(A ∩ B) on one side to P(A ∩ B) ≥ P(A) + P(B) − 1.

Law of total Probability

P(A) = sigma[P(A|Bn)P(Bn)]

2.1.7 you roll a die, and i roll a die. You win if the number showing on your die is strickly greater than the one on mine. If we play this game fice times, what is the chance that you win at least four times?

The probability that you win in one game is half of the chance to NOT have a double (by symmetry). This probability is, therefore,p=1/2×(1−1/6) =5/12.The answer to the problem now is the probability that the number of successes out of n= 5 trials (with probability of successp=5/12) is at least 4. This is equal to(5 4)(5/12)^4(7/12)+(5 5)(5/12)^5= 5×(5/12)^4(7/12)+(5/12)^5= 0.1005.

2.1.5 Given there were 12 heads in 20 independent coin tosses,calculate a) the chance that the first toss is landed heads; b) the chance that the first two tosses landed heads; c) the chance that at leas two of the first five tosses landedheads

just look at the answers

2.4.5 Suppose that each week you buy a ticket in a lottery which gives you a chance of 1/100 of a win. You do this each week for a year. What is the chance that you get k wins during the year approximately? Calculate as a decimal for k = 0,1,2.

we have n = 52 weeks in a year and the probability p= 1/100 to win the lottery every single week. Approximating with the Poisson distribution for μ=np=52/100 we obtain that, the approximate probability to have k wins every year is e^-52/100^(52/100)^k/k!. k= 0 :59.45% k= 1 :30.92% k= 2 :8.03% .

1.4.9 Three high schools have senior classes of size 100,400, and 500, respectively. Here are two schemes for selecting a student from among the three senior classes: A: Make a list of all 1000 seniors, and choose a student at random from this list. B: Pick one school at random, then pick a student at random from the senior class in that school. Show that these two schemes are not probabilistically equivalent. here is a third scheme: C: Pick school i with probability pi (p1+p2+p3 =1), then pick a student at random from the senior class in that school. Find the probabilities p1, p2, and p3, which make scheme C equivalent to scheme A.

9 Using scheme A, any student (regardless of the school they are attending) has a 1 in 1000 chance to be picked. Using scheme B, a student in school 1 has a 1/3 × 1/100 != 1/1000 to be picked up. The two schemes are not equivalent. In scheme C, 1. any student in school 1 has a p1 1/100 chance 2. any student in school 1 has a p2 1/400 3. any student in school 1 has a p3 1/500 In order to make scheme C equivalent to scheme A we need (p1 1/100) = (p2 1/400) = (p3 1/500) = 1/1000 , so p1 = 0.1, p2 = 0.4, p3 = 0.5.

2.2.13 A pollster wishes to know the percentage of p of people in a population who intend to vote for a particular candidate. How large must a random sample with replacement be in order to be at least 95% sure that the sample percentage is within one percentage point of p?

Finding confidence intervals is based on the very important computation P(|pˆ− p| ≤ l √pq /√ n ) = Φ(l) − Φ(−l), where the above equality is approximate. With the crucial observation that √pq ≤ 1/2 for any value of p, this leads to the confidence interval formula P(|pˆ− p| ≤ l/ 2 √ n ) ≥ Φ(l) − Φ(−l). 2 In this particular problem, we need the confidence level Φ(l) − Φ(−l) = 95%. This amounts (from the table ) to l = 1.96 (note it is a bit lower than 2). Translating the words into mathematics, we actually need an n large enough such that l /2 √ n ≤ 1/ 100 for l = 1.95. This means that n ≥ ( 196/ 2)^2 = 9604.

1.3.15 Use Boole's inequality and the fact that (U[n i=1] Ai) = ∩ [n i=1 A^ci to show that P(B1B2...Bn) >= sigma P(Bi) - (n-1)

First, from the principle of inclusion-exclusion P(A ∪ B) = P(A) + P(B) − P(A ∩ B), we make the (rather obvious) observation that P(A ∪ B) ≤ P(A) + P(B), because P(A ∩ B) ≥ 0. So, "probability of the union is at most the sum of probabilities". This statement generalizes easily to more than two events (by induction, for example).

2.2.5 Suppose you bet a dollar on red, 25 times in a row, at roulette. Each time you win a dollar with probability 18/38, lose with probability 20/38. Find, approximately, the chance that after 25 bets you have at least as much money as you started with.

In order to be ahead in a bet involving 1 to 1 money odds, one has to win more plays than one loses. Out of n = 25, this means that the player needs to win at least a = 13. Consequently, we need to compute/approximate the probability for n = 25, p = 18/38 , a = 13, b = ∞, by by 1 − Φ ( a − 1/ 2 − np/ √npq ) = 1 − Φ(0.26) = 0.5 − 0.1026 = 0.3974.

1.4.4 Two events have probability of 0.1 and 0.3. What is the probability that: a) neither of the events occur? b) at least one of the events occurs? c) exactly one of the events occurs?

Let us name the two events A and B. We know that they are independent and that P(A) = 0.1 and P(B) = 0.3. We now compute the probabilities of a) neither of the two events occur, which means AcBc . Since A and B are independent, so are their complements, so P(AcBc) = P(Ac)P(Bc) = (1 − 0.1)(1 − 0.3) = 0.9 × 0.7 = 0.63. b) at least one of the events occur, which is in set notation A ∪ B. The probability is, according to the principle of inclusion-exclusion P(A ∪ B) = P(A) + P(B) − P(AB) = 0.1 + 0.3 − 0.1 × 0.3 = 0.37, where we used independence to compute P(AB) as the product of probabilities. c) exactly one of the events occur, which means ABc ∪ AcB. The probability is (sincethe union is actually a partition) and using independence 0.1 × 0.7 + 0.9 × 0.3 = 0.34.

3.2.11 There are 100 prize tickets among 1000 tickets in a lottery. What is the expected number of prize tickets you will get if you buy 3 tickets? What is a simple upper bound for the probability that you will win at least one prize? Compare with the actual probability. Why is the bound so close?

Markov's inequality = P(X >= a) <= E[x]/a The probability that some random variable x is greater than or equal to some constant a, is less than or equal to the expected value of x divided by constant a .

1.6.1 There are twelve signs of the zodiac. How many people must be present for there to be a 50% chance that two or more of them were born under the same sign?

P(2 or more same) >= .5 P(2 or more) = 1- P(n all different) P(n different) = (12-i/12)

1.3.11 Inclusion-exclusion formula for 3 events. Write AuBuC = (AuB)uC and use the inclusion-exclusion formula three times to derive the inclusion-exclusion formula for 3 events: P(AuBuC) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC)

P(A ∪ B ∪ C) = P(A ∪ (B ∪ C)) = P(A) + P(B ∪ C) − P(A ∩ (B ∪ C)) = P(A) + P(B ∪ C) − P((A ∩ B) ∪ (A ∩ C)) = P(A) + [P(B) + P(C) − P(B ∩ C)] − [P(A ∩ B) + P(A ∩ C) − P((A ∩ B) ∩ (A ∩ C))]= P(A) + P(B) + P(C) − P(A ∩ B) + P(B ∩ C) − P(C ∩ A) + P(A ∩ B ∩ C)).

1.4.12 Give a formula for P(F|Gc) in terms of P(F), P(G), and P(FG) only.

P(F/Gc) = P(F Gc)/P(Gc) = (P(F) − P(F G))/ (1 − P(G)).

Use consecutive odds ratios to find the largest k that maximizes the Poisson (mew) probability Pmew(k). For what values of mew is there a double maximum? What are the two values of k in that case? Is there ever a triple maximum?

To do the consecutive odds ratios just put the poisson approximations on top of each others, with the bottom one with (k-1) which equal mew/k. Therefore, the probability P(k) compares to the probability P (k−1) exactly the same way as μ compares to k . Very similarly to the way we found the mode for the binomial distributions,this means that we actually have two cases: 1. if μ >0 is NOT an integer then the mode is the preceding integer to μ , denoted [μ]. 2. ifμ > 0 is an integer, then we have two modes μ and μ −1 (as we note that for k = μ we have P ( k) = P(k−1).) There is NO other situation, which means there is no possibility to have a triple mode. To have a triple mode we we need to have the equality μ = k for more that one possible integer value of k. This is clearly not possible.

2.1.8 For each positive integer n, what is the largest value of p such that zero is the most likely number of successes in n independent trials with success probability p?

We are looking for the largest p such that the integer part of (n+ 1)pis equal to zero.This is equivalent to (n+ 1)p <1. The largest value for p is, therefore, 1/(n+ 1). In the case of equality,p= 1/(n+ 1), we have two modes, at k= 1 and k= 0.

1.4.5 There are two urns. The first urn contains 2 black balls and 3 white balls. The second urn contains 4 black balls and 3 white balls. An urn is chosen at random, and a ball is chosen at random from that urn. a) Draw a suitable tree diagram b) Assign probabilities and conditional probabilities to the branches of the tree. c) Calculate the probabilities that the ball drawn is black.

We denote by U1 and U2 the events that urn 1 or urn 2 are picked, and by B and W the events that a white ball or a black ball is picked. b) the probabilities on the first "branches" are P(U1) = P(U2) = 1/2 2. The conditional probabilities on the next "branches" of the tree are P(B/U1) = 2/5, P(W/U1) = 3/5, P(B/U2) = 4/7, P(W/U2) = 3/7. c) the law of total probability says P(B) = P(U1)P(B/U1) + P(U2)P(B/U2) = 1/2×2/5+1/2×4/7=34/70= 17/35.

2.1.11 70% of the people in a certain population are adults. A random sample of size 15 will be drawn, with replacement, from this population. a) What is the most likely number of adults in the sample? b) What is the chance of getting exactly this many adults?

We haven= 15 andp= 0.7. a) To find the mode, we see that np+p= (n+ 1)p= 16×0.7 = 11.2. Therefore, themode (the most likely number of successes), ism= 11. b) Now, for these n, k, we have P(11) =(15 11)×0.7^11(0.3)^4= 1365×0.00016016346 = 0.218623.

2.1.3 Suppose 5 dice are rolled. Assume they are fair and the rolls are independent. Calculate the probability of the following events: A= (exactly two sixes);B= (at least two sixes);C= (at most two sixes);D= (exactly three dice show 4 or greater);E= (at least 3 dice show 4 or greater).

a) P(2) =(5 2)(1/6)^2(5/6)^3=[(4×5)/(1×2)]×[125/7776]= 0.16075. b) The answer is P(2) +P(3) +P(4) +P(5) = 1−P(0)−P(1) = 1−(5/6)^5+ 5×1/6(5/6)^4= 1−31257776−31257776=15267776= 0.19624. c) Using the above computations, we have P(C) =P(0) +P(1) +P(2) = (1−0.19624) + 0.16075 = 0.96451. d) and e), we havep= 1/2, so: d)P(3) =(5 3)(1/2)^5=10/32= 0.3125. e) here we can use symmetry to obtainP(E) =P(3) +P(4) +P(5) =P(0) +P(1) +P(2) = 0.5.

1.1.4 Equally Likely Outcomes Suppose I bet on red at roulette and you bet on black, both bets on the same spin of the wheel. a) What is the probability that we both lose? b) What is the probability that at least one of us wins? c) What is the probability that at least one of us loses?

a) 2/38 because only 2 options out of 38 are neither black or red. b) 36/38 this is the opposite of a c) 100% at least 1 player always loses.

1.1.8 Repeat exercise 7 for two rolls of a n-sided die for an arbitrary n instead of 6. a) The maximum of the two numbers rolled is less than or equal to 2 b) maximum of the two numbers rolled is less than or equal to 3. c) the maximum of the two numbers is exactly equal to 3 d) repeat b & c for x instead of 3, for each x from one to 6 e) Denote P(x) the probability that the maximum number is exactly x. What should P1+p2+p3+p4+p5+p6 equal?

a) 4/n^2 b) 9/n^2 c) 5/n^2 d) (x^2/n^2) , (2x-1)/n^2 e) n^2/n^2 = 1

1.1.5 Suppose a deck of 52 cards is shuffled and the top two cards are dealt. a) How many ordered pairs of cards could result in an outcome? b) Assuming each of the pairs has an equal chance, calculate the chance the first card is an ace c) the chance the second card is an ace (explain your answer by symmetry argument as well as by counting) d) the chance that both cards are aces e) the probability that at least one is an ace

a) 52 x 51 = 2652, you have to subtract by one the second time because the cards are not replaced. b) 4x51 = 204 pairs with first card ace, then 204/2652 = 1/13 c) because the outcomes are equally likely, it doesn't matter if it comes first or second. Therefore by the symmetry rule it is exactly the same as if it were first 1/13. d) 4x3= 12 pairs both aces. 12/2652 = 1/221 e) P(first card ace) +P(second card ace) - P(intersection) 1/13 + 1/13 -1/221 = 33/221

1.1.6 Repeat exercise 5, supposing instead that after the first card is dealt, it is replaced, and shuffled into the deck before the second card is dealt. a) How many ordered pairs of cards could result in an outcome? b) Assuming each of the pairs has an equal chance, calculate the chance the first card is an ace c) the chance the second card is an ace (explain your answer by symmetry argument as well as by counting) d) the chance that both cards are aces e) the probability that at least one is an ace

a) 52 x 52 =2704 b) 4x52 same c) 4x52 same d) 4x4 = 16, 16/2704=1/169 e) 1/13 + 1/13 -1/169 = 25/169

Suppose that the birthday of each of three people is equally likely to be any one of the 365 days of the year, independently of others. Let B(sub ij) denote the event that person i has the same birthday as person j, where the labels i and j may be 1, 2, or 3. a) Are the events B(sub 12) and B(sub 23) independent? b) Are the events B(sub 12), B(sub 23), and B(sub 13) independent? C) Are the events B(sub 12), B(sub 23), and B(sub 13) pairwise independent?

a) Clearly,P(B12) =P(B23) =1/365.This is just the probability that one additional person has the same birthday as another (the previous) one. b) The events B12B23=B12B23B31 both mean all three people have the same birthday.The probability of this can be easily computed, asking the second to have the same birthday as the first, and then the third still have the very same birthday as the previous two: P(B12B23) =P(B12B23B31) =13652. This means that P(B12B23B31)! =P(B12)P(B23)P(B31) =13653. The three events are NOT independent. c) according to the computation in a) and b) we haveP(B12B23)=13652=P(B12)P(B23) so B12 and B23are independent. Similarly, any other pair of the three events are independent,so they are pair-wise independent.

1.1.7 Suppose two dice are rolled. Find the probability of the following events. a) The maximum of the two numbers rolled is less than or equal to 2 b) maximum of the two numbers rolled is less than or equal to 3. c) the maximum of the two numbers is exactly equal to 3 d) repeat b & c for x instead of 3, for each x from one to 6 e) Denote P(x) the probability that the maximum number is exactly x. What should P1+p2+p3+p4+p5+p6 equal?

a) P(number rolled is <= 2) = (2x2)/(6x6) = 1/9, There are only two options for first and second dice, and you find the total options by squaring the options of one. b) P(number rolled is <= 3) (3x3)/36 = 1/4 c) P(the maximum rolled is exactly 3) (1,3) (3,1) (2,3) (3,2) (3,3) =5/36 d) P(max number rolled is <=x) = x^2/36 P(max number rolled is exactly x) = (2x-1)/36, for this one draw a table e. (1/36) + (3/36) + (5/36) + (7/36) + (9/36) + (11/36) = 1

1.5.6 An experimenter observes the occurrence of an event A as the result of a particular experiment. There are three different hypotheses, H1, H2, and H3, which the experimenter regards as the only possible explanations of the occurrence of A. Under hypothesis H1, the experiment should produce the result A about 10% of the time over the long run, under H2 about 1% of the time, and under H3 about 39% of the time Having observed A, the experimenter decides that H3 is the most likely explanation, and that the probability that H3 is true is (39%)/(10%+1%+39%) = 78% a) What assumption is the experimenter implicitly making? b) Does the probability 78% admit a long-run frequency interpretation? c) Suppose the experiment is a lab test on a blood sample from an individual chosen at random from a particular population. The hypothesis Hi is that the individual's blood is of some particular type i. Over the whole population it is known that the proportion of individuals with blood of type 1 is 50%, the proportion with type 2 blood is 45%, and the remaining proportion is type 3. Revise the experimenter's calculation of the probability of H3 given A, so that it admits a long-run frequency interpretation. Is H3 still the most likely hypothesis given A?

a) The experimenter is making the implicit assumption that the three hypotheses are equally likely (prior to the experiment). b) No, because the implicit assumption above is a subjective belief of the experimenter. c) We compute the conditional probabilities: P(H1/A) = (.5 × .1) /(.5 × .1 + .45 × .01 + .05 × .39), P(H2/A) = (.45 × .01)/ (.5 × .1 + .45 × .01 + .05 × .39), P(H2/A) = (.05 × .39)/(.5 × .1 + .45 × .01 + .05 × .39). With these new priors, the most likely cause after the observation is H1 (largest denumerator)

1.5.3 A manufactoring process produces integrated circuit chips. Over the long run the fraction of bad chips produced by the process is around 20%. Thoroughly testing a chip to determine whether it is good or bad is rather expensive, so a cheap test is tried. All good chips will pass the cheap test, but so will 10% of the bad chips. a) Given a chip passes the cheap test, what is the probability that it is a good chip? b) If a company using this manufacturing process sells all chips which pass the cheap test, over the long run what percentage of chips sold will be bad?

a) This is clearly Bayes' rule. Denote by B and G the events that the chip is bad or good, and by PS the event that a chip passes the test. The assumptions are that P(B) = .2, P(G) = .8, P(PS/G) = 1, P(PS/B) = .1. We are asked to compute P(G/PS) = (.8 × 1 )/(.8 × 1 + .2 × .1)= 8/8.2 = 40/41. b) the long run percentage of bad chips who passed the test is equal to P(B/PS) = 1 − P(G/PS) = 1/41.

2.4.7 . Let S be the number of successes in 25 independent trials with probability 1/10 of success on each trial. Let m be the most likely value of S. a) Find m. b) Find P(S = m) correct to three decimal places. c) Compute the normal approximation to P(S = m). d) What is the value of the Poisson approximation to P(S = m)? e) Repeat parts (a), (b), (c) and (d) for n = 2500 instead of 25, and determine which approximation, the normal or the Poisson, is better. f) Repeat part (e) for n = 2500 and p = 1/1000 instead of p = 1/10.

a) We know thatthe mode m for the binomial distribution is the preceding integer value to (n+ 1)p = (25 + 1)/10 = 2.6, so m = 2. b) the exact probability is P(2) =(25×24/1×2)0.1^2 x 0.9^23 = .2659 c) normal approximation use the regular one Φ(0)−Φ(−2/3) = Φ(2/3)−Φ(0) = 0.2475 d) use poisson e^-2.5^(2.5)^2/2! = 0.2565 e) for n= 2500, we have that σ=√npq = 153 This is significantly larger than 3, so we use "the rule of thumb" to say that the normal approximation is the better one. The mode is m = 250. Now, compute ) Now, if p= 1/1000 then np = 2.5 is rather small, to the extent that σ≈√2.5 is much smaller than 3. The Poisson approximation is better (p is small enough compared to n). Again, the mode for this binomial distribution is m= 2. Using Poisson approximation, we have, AGAIN

1.3.4 Let Ω = {0,1,2} be the outcome space in a model for tossing a coin twice and observing the total number of heads. Say if the following events can be represented as subsets of Ω. If you say "yes", provide the subset; if you say "no", explain why: a) the coin does not land heads both times; b) on one of the tosses the coin lands heads, and on the other toss it lands tails c) on the first toss the coin lands heads, and on the second toss it lands tails d) the coin lands heads at least once.

a) Yes. "The coin does not (land heads both times)" means that there are either 0 or 1 H's. This event can be written as the subset {0, 1} of Ω. b) Yes. This means there is exactly one H out of the two tosses. The subset is {1} ⊂ Ω. c) No, this is cannot be represented as a subset of Ω. Just knowing the total number of H's we cannot distinguish in between HT and T H. In other words, the total number of H's does not keep track of the order. d) Yes. The subset is {1, 2} ⊂ Ω.

2.2.1 let H be the number of heads in 400 tosses of a fair coin. Find normal approximations to: a) P(190 <= H <= 210) b) P(210 <= H <= 220) c) p(H=200) b) P(H=210)

a) a=190, b=210 Φ(1.05) − Φ(−1.05). If we use, for example, the table provided on Canvas (and in class), the area above represents exactly twice the area in the table, which means Φ(1.05) − Φ(−1.05) = 2 × .3531 = .7062 b) here a = 210, b = 220. The approximate probability is Φ (20.5 /10 ) −Φ ( 9.5/ 10 ) = Φ(2.05)−Φ(0.95) = (1 2 +.4798)−( 1 2 +.3289) = .4798−.3289 = .1509 c) a = b = 200. Note that, without the continuity correction, the approximation would be zero, which is not meaningful. With the continuity correction, the approximation is Φ ( 0.5/ 10) − Φ (− 0.5/ 10 ) = Φ(0.05) − Φ(−0.05) = 2 × 0.0199 = 0.0398 d) a = b = 210. Φ ( 10.5 /10) − Φ ( − 9.5 /10 ) = Φ(1.05) − Φ(.95) = .( 1 2 + 0.3531) − ( 1 2 + 0.3289) = 0.0242.

1.2.3 (Explain) Suppose there are 10 horses in a race and a bookmaker quotes odds of ri to 1 against horsei winning. Let pi = 1/ri +!, i = 1 to 10, so each pi is between 0 and 1. Let sigma p1 +....p10 a) Do you expect that the sum is greater than, smaller than, or equal to 1? Why? b) suppose the sum were less than 1. Could you take advantage of this? How? [Hint: by betting on all 10 horses in the race, a bettor can win a constant amount of money, regardless which horse wins.

a) because the bookmaker always wants to win he will inflate his odds. 1/(ri+1) > 1/(rfair +1). Therefore sigma pi >1 b) if sigma is less than 1, then we can make money off of betting on this situation. If we bet $x, such that $x = pi/sigma, then if we bet the same amount on each horse, we are guaranteed to make money. (go back and look later)

2.1.12 A gambler decides to keep betting on red at roulette, and stop assoon as she has won a total of five bets. a) what is the probability that she has to make exactly 8 betsbefore stopping? b) what is the probability that she has to make at least 9 bets?

a) exactly 8 bets before stopping means exactly 4 reds out of the first 7 bets, then a red in the 8th bet. This happens with probability (7 4)(18/38)^4(20/38)^3×18/38. b) at least nine bets to the 5th red is the same as the first 8 bets show at most 4 reds.This probability is 4∑k=0(8 k)(18/38)^k(20/38)^8−k.

1.5.4 A digital communications system consists of transmitter and a receiver. During each short transmission interval the transmitter sends a signal which is to be interpreted as A zero, or it sends a different signal which is to be interpreted as a one. At the end of each interval, the receiver makes its best guess at what was transmitted. Consider the events: 𝑇0 = {transmitter sends 0}, 𝑅0 = {receiver concludes that a 0 was sent} 𝑇1 = {transmitter sends 1}, 𝑅1 = {receiver concludes that a 1 was sent} Assume that: 𝖯(𝑅0∣𝑇0)=0.99,𝖯(𝑅1∣𝑇1)=0.98,𝖯(𝑇1)=0.5 Find: (a) the probability of A transmission error given 𝑅1; (b) the overall probability of a transmission error; (c) Repeat (a) and (b) assuming 𝖯(𝑇1)=0.8 instead of 0.5

a) given R1 a transmission error means that a 0 was sent. We are asked to computethe probability P(T0/R1) = (1/2 × .01)/(1/2× .01 + 12 × .98) = 1/99 b) probability of transmission error is P(T0R1) + P(T1R0) = (1/2× .01 ) +(1/2× .02 )= .015 c) P(T0/R1) = (.2 × .01)/(.2 × .01 + .8 × .98) = .0025 P(T0R1) + P(T1R0) = .2 × .01 + .8 × .02 = .018

2.2.3 A fair coin is tossed repeatedly. Consider the following two possible outcomes: 55 or more heads in the first 100 tosses 220 or more heads in the first 400 tosses a) Without calculation, say which of these outcomes is more likely. Why? b) Confirm your answer to a) by a calculation

a) the larger the n the smaller the probability. b) look it up

1.3.10 Events A,B, and C are define in an outcome space. Find expressions for the following probabilities in terms of P(A), P(B), P(C), P(AB), P(AC), P(BC), P(ABC). a) The probability that exactly two of A,B,C occur. b) The probability that exactly one of these events occurs. c) The probability that none of these events occur.

a) the probability that exactly two events occur can be written as: P(ABCc) + P(BCAc) + P(CABc). This is equal to (P(AB) − P(ABC)) + (P(BC) − P(ABC)) + (P(CA) − P(ABC)) = = P(AB) + P(BC) + P(CA) − 3P(ABC). b) the probability that exactly one of the events occurs is P(ABcCc) + P(BCcAc) + P(CAcBc). Let us work on one term (as the other are obtained by re-shuffling the symbols). P(ABcCc) = P(A ∩ (B ∪ C)c) = P(A) − P(A ∩ (B ∪ C)) = P(A) − P((A ∩ B) ∪ (A ∩ C)). Using the principle of inclusion-exclusion, we obtain P(ABcCc) = P(A) − P(AB) − P(AC) + P(ABC). We permute the symbols to obtain similar expressions for P(BCcAc) = P(B) − P(BC) − P(BA) +P(ABC) and P(CAcBc) = P(C) − P(CA) − P(CB) + P(ABC). Upon addition of the three, we get P(A) + P(B) + P(C) − 2 × (P(AB) + P(BC) + P(CA)) + 3P(ABC). c) passing to the complement we have that the probability of none occurring is 1 − P(A ∪ B ∪ C) = 1 − P(A) − P(B) − P(C) + P(AB) + P(BC) + P(CA) − P(ABC). We have used above the principle of inclusion-exclusion for three events, namely problem

1.5.9 A box contains three "shape" as described in Example 1.3.3. one of the shapes is a fir die, and lands flat with probability 1/3. The other two shapes land flat with probabilities 1/2 and 2/3, respectively. a) One of the three shapes will b e chosen at random, and rolled. What is the chance that the number rolled is 6? b) Given that the number rolled is 6, what is the chance that the fair die was chosen?

a) this is the law of total probability, computed as (1/3)(1/6)+(1/3)(1/4)+(1/3)(1/3)= 9/36 = 1/4 b) we can use Bayes (the denominator of the fraction is computed above) to obtain P(f air die/given 6) = ((1/3)(1/6))/(9/36) = 2/9

a typical slot machine in a Nevada casino has 3 wheels, each marked with 20 symbols at equal spacings around the wheel. the machine is engineered so that on each play the 3 wheels spin independently, and each wheel is equally likely to show any one of the 20 symbols when it stops spinning. on the central wheel, 9 of the 20 symbols are bells, while there is only one bell on the left wheel and one on the right wheel. the machine pays out the jackpot only if the wheels come to rest with each wheel showing a bell. a) calculate the probability of hitting the jackpot. b) calculate the probability of getting 2 bells but not the jackpot. c) suppose that instead there were 3 bells on the left, 1 on the middle, and 3 on the right. how would this affect the probabilities in a) and b)? explain why the casino might find the 1-9-1 machine more profitable than a 3-1-3 machine.

a) three bells (on all wheels at the same time) show with probability 1/20×9/20×1/20=9/8000= 0.001125. b) We have three separate situations, where the "no bell" shows up on the left wheel,middle or right wheel. Altogether, it adds up to 19/20×9/20×1/20+1/20×11/20×1/20+1/20×9/20×19/20=3538000= 0.0441.

1.6.3 A biased coin lands heads with probability 2/3. The coin is tossed 3 times a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads? b) use your answer in a) to find the probability that there was exactly one head, give that there was at least one head in the three tosses.

a) using the definition of conditional probability, and the fact that at least two head simplies at least one head, the probability we are looking for is (one way to write it, since we did not yet learn the binomial distribution here) P{HHH, T HH, HT H, HHT}/ 1−P({T T T})= [3×(2/3)^21/3 + (2/3)^3]/[1−(1/3)^3]=20/26= 0.7692. b) conditioned on having at least one H, we can either have exactly one H or at least 2 H's. In other words, the conditional probability here is the complement to the conditional probability found in part a). The answer is 1−0.7692 = 0.2308.

2.2.9 An airline knows that over the long run 90% of passengers who reserve seats show up for their flight. On a particular flight with 300 seats, the airline accepts 324 reservations. a) Assuming that passengers show up independently, what is the chance that the plane will be overbooked? Give answer as a decimal to four correct decimals. b) Suppose that people tend to travel in groups. Would that decrease or increase the probability of overbooking? Explain your answer. c) Redo the calc a) assuming that passengers always travel in pairs. Check that your answers to a), b), and c) are consistent

a) we need to to the approximation for a = 301, n = 324 = 0.9 with b = ∞. This comes to (already done this homework) 1 − Φ ( a − 1/ 2 − np/ √npq ) = 1 − Φ(1.648)) = .5 − .4505 = 0.0495 b) similarly to problem 3 a), a smaller n (grouping people) will tend to increase the probability of overbooking. c) we have to do the computations for n = 324/2 = 162, p = .9, a = 150 + 1 = 151. We obtain, exactly as in part a) 1 − Φ ( a − 1/ 2 − np/ √npq ) = 1 − Φ(1.23) = .5 − .3907 = 0.1093.

1.3.9 Events F, G, and H are such that P(F) = 0.7, P(G) = 0.6, F(H) = 0.5, P(FG) = 0.4, P(FH) = 0.3, P(GH) = 0.2, P(FGH) = 0.1 Find: a) P(F u G) b) P(F u G u H) c) P(F^c G^c H)

a) we use the principle of inclusion-exclusion: P(F ∪ G) = 0.7 + 0.6 − 0.4 = 0.9 b) we use the inclusion-exclusion principle for three events (problem 1.3.11 below) P(F ∪ G ∪ H) = 0.7 + 0.6 + 0.5 − 0.4 − 0.3 − 0.2 + 0.1 = 1. c) P(FcGcH) = P(H ∩ (F ∪ G)c) = P(H) − P(H ∩ (F ∪ G)) =P(H) −P((H ∩ F) ∪(H ∩G)) Using principle of inclusion-exclusion for the last probability above we obtain 0.5 − (0.3 + 0.2 − 0.1) = 0.1