Probability Analysis of Binomial Random Variables and Hypothesis Testing, Assignments of Statistics

Download Probability Analysis of Binomial Random Variables and Hypothesis Testing and more Assignments Statistics in PDF only on Docsity! University Problem Set #6: Solutions ECE 313 of I l l inois Page 1 of 4 Spring 2002 1.(a) X is a binomial random variable with parameters (10, 0.5) and mean 10•0.5 = 5. (b) P{X ≥ 4} = 1 – P{X ≤ 3} = 1 – 2–10[ ]1 + ( )101 + ( )102 + ( )103 = 1 – 1761024 = 8481024 = 5364. P{X ≤ 5 | X ≥ 4} = P{4 ≤ X ≤ 5} P{X ≥ 4} = 2 –10[ ] ( )104 + ( )105 /P{X ≥ 4} = 4621024×1024848 = 231424. (c) P{4th toss = Head | X = 4} = P{4th toss = Head and X = 4} P{X = 4} = P{4th toss = Head and 3 Heads in other 9 tosses} P{X = 4} = P{4th toss = Head}P{3 Heads in other 9 tosses} P{X = 4} (by independence of tosses) = (1/2)( )93 2–9 ( )104 2–10 = 4 10 . (d) For any arbitrary value of P{Heads} = p > 0, we have that P{4th toss = Head | X = 4} = P{4th toss = Head}P{3 Heads in other 9 tosses} P{X = 4} = p( )93 p3(1–p)6 ( )104 p4(1–p)6 = 4 10 . Thus, not knowing p does not disadvantage me; the probability is 4/10 regardless of the value of p. Now, a fair bet should be offering 3-to-2 odds, i.e. with a bet of $1, you win $1.50 roughly 40% of the time and lose $1 roughly 60% of the time. A bookie who offered 2-to-1 odds would be losing $0.20 per dollar bet and would soon be out of business, and perhaps wearing concrete overshoes as well! The fact that he knows the outcome of the 4th toss and yet is offering such great odds to induce you to bet that a Head occurred, leads to the suspicion that the only reason the bookie can afford to offer these odds is that the 4th toss resulted in a Tail, and thus is sure that he is not going to lose. I would not bet on a Head. 2.(a) Obviously United Airlines has better on-time performance at all five airports. (b) P(T|UC)P(C|U) = P(TUC) P(UC) × P(UC) P(U) = P(TUC) P(U) and similarly for the other terms. Hence, the right side of the given expression is P(TUC) + P(TUL) + P(TUX) + P(TUD) + P(TUF) P(U) . But, (TUC) ∪ (TUL) ∪ (TUX) ∪(TUD) ∪(TUF) is a partition of the set TU, and thus the numerator above is just P(TU), and the ratio is, by definition, P(T|U). Also, P(T|W) = P(T|WC)P(C|W)+P(T|WL)P(L|W)+P(T|WX)P(X|W)+P(T|WD) P(D|W)+P(T|WF)P(F|W) (c) Plugging and chugging, P(T|U) = 0.8655 < 0.896 = P(T|W). The reason for the discrepancy is that America West has most of its flights into sunny Phoenix where flights have a good chance of being on time. In contrast, United has only a few flights to Phoenix and has lots of flights into snowy Chicago (and foggy San Francisco.) Note that 0.69 of the 0.896 of P(T|W) comes from Phoenix, whereas United gets only 0.0475 of the 0.8655 of P(T|U) from Phoenix (where it performs the best). The moral for the marketing departments is that America West should advertise itself as most-often-on-time, whereas United should advertise itself as most-often-on-time-where-you-wanna-go. Comment: This problem is based on a real-life case. The University of California at Berkeley (of all places!) was accused by the EEOC of discriminating against women applying to graduate school. Each Department actually admitted a larger fraction of its women applicants than of its men applicants, but the overall statistics showed a smaller fraction of women applicants being admitted as compared to men applicants. 3 . Let A denote the event that a baby survives delivery and B the event that it is delivered by C section. Then, we are given that P(A) = 0.98, P(A|B) = 0.96, and P(B) = 0.15. The theorem of total probability tells us that P(A) = 0.98 = P(A|B)P(B) + P(A|Bc)P(Bc) = 0.96×0.15 + P(A|Bc)×0.85 ⇒ P(A|Bc) ≈ 0.9835. Quick sanity check: P(A|B) < P(A) < P(A|Bc), so we haven’t made an obvious calculation error. 4 . There are ( )112 = 11×101×2 = 55 pairs of letters that might fall off. The 55 pairs can be classified into the pair OO, 5 other pairs from the set {H, O, O, N}, 4×7 = 28 pairs consisting of one letter from {H, O, O, N} and one from {C, A, T, T, A, G, A}, 3 pairs AA and 1 pair TT, and 17 other pairs from {C, A, T, T, A, G, A}. If OO fell off, the sign will always read correctly. If one of 5 other pairs from {H, O, O, N} fell off, the letters will always look right side up, but will be interchanged with probability 1/2. For the 28 pairs consisting of one letter from {H, O, O, N} and one from {C, A, T, T, A, G, A}, the restored sign is correct with probability 1/4, incorrect but readable with probability 1/4, and one letter University Problem Set #6: Solutions ECE 313 of I l l inois Page 2 of 4 Spring 2002 appears inverted with probability 1/2. For the AA and TT pairs, the sign is correct with probability 1/4, has one letter inverted with probability 1/2 and has both letters inverted with probability 1/4. The remaining 17 pairs are put back correctly with probability 1/8, interchanged but upright with probability 1/8, have one letter upside down with probability 1/2, and have both letters upside down with probability 1/4. This gives (a) P(sign reads CHATTANOOGA) = [ ]1 + 5×12 + 28×14 + 4 ×14 + 17–18 × 155 = 109440 ≈ 25%. Not Bad! (b) P(sign readable but not CHATTANOOGA) = [ ]5×12 + 28×14 + 17×18 × 155 = 93440 (c) P(one letter seems to be upside down) = [ ]28×12 + 4 ×12 + 17×12 × 155 = 196440 = 49110 (d) P(two letters seem to be upside down) = [ ]4×14 + 17×14 × 155 = 42440 = 21220 Sanity check: The sum of the probabilities is (109+93+196+42)/440 = 1. (e) All letters seem to be right side up includes the case when the sign reads CHATTANOOGA! Now, P(at least one vowel|all seem to be right side up) = P(all seem to be right side up∩ at least one vowel) P(all seem to be right side up) We know from (a) and (b) that the denominator is (109+93)/440 = 101/220. Now, our list above can be further classified into 1 pair OO and 2 pairs each of the forms HO and NO (for these the letters always appear to be right side up), 14 of the form O and one from {CATTAGA} and 3 each of the form HA and NA (for these, the probability that the letters seem right side up is 1/2), 3 of the form AA, and 3 each of the form CA and GA, and 6 of the form TA (for these the probability that tbe letters seem to be right side up is 1/4). All other pairs {from the set CHTTNG} do not include a vowel. Check: P(at least one vowel) = 1 – P(no vowel) = 1 – [(6×5)/(1×2)]/55 = 40/55 = (1 + 2 + 2 + 14 + 3 + 3 + 3 + 3 + 3 + 6)/55 so we got 'em all! From all this, we can readily compute P(all seem to be right side up ∩ at least one vowel) = [(1+2+2)×1 + (14 + 3 + 3)×1/2 + (3 + 3 + 3 + 6)×(1/4)]/55 = 75/220. Hence, P(at least one vowel | all seem to be right side up) = 75 220 / 101 220 = 75 101 . (f) The designated driver can correctly identify the letters that fell down if and only if the driver can see that two letters are obviously switched (possibly being turned upside down) or if two letters both appear to be upside down but in their correct places. Note that these are disjoint possibilities. For the various pairs of different types, we see that if OO fell down, the sign will still read CHATTANOOGA and the driver will not be able to identify with certainty which letters fell down.. If one of the 5 other pairs from {H, O, O, N} fell down, the driver can identify them if and only if they have been switched (probability 1/2); whether they have been flipped over doesn’t matter. For the 28 pairs consisting of one letter from {H, O, O, N} and one from {C, A, T, T, A, G, A}, the two letters must be swapped in order to be identifiable (probability 1/2), while for the 3 + 1 pairs AA and TT, both letters must be upside down to be identifiable with certainty. Finally, the remaining 17 other pairs from {C, A, T, T, A, G, A}, either both letters must be in the wrong place (probability 1/2) or they must be in the right place and both upside down (probability 1/8). Hence, P{driver can identify correctly} = [ ]155×0 + 555×12 + 2855×12 + 455×14 + 1755×58 = 225440 = 4588 > 50%!! (g) Since all pairs are equally likely to fall down, the Bayes’ decision is the same as the maximum-likelihood decision. Since P(CHATTANOOGA | OO fell down) = 1 and P(CHATTANOOGA | XY fell down) < 1 for any other choices of X and Y, the Bayes' (and maximum likelihood) decision favors OO falling down whenever the sign reads CHATTANOOGA. From part (a), the conditional probability that the decision is correct when the sign reads CHATTANOOGA is P(OO | CHATTANOOGA) = P(CHATTANOOGA | OO)P(OO)/P(CHATTANOOGA) = 1×(1/55)/(109/440) = 8/109. 5.(a) Since the pitcher can only throw fastballs, curveballs or sliders, P(F) + P(C) + P(S) = 1. Also, P(C) = 2P(F) so we conclude that P(S) = 1 – 3P(F). Now, 1/4 = P(H|C)P(C) + P(H|F)P(F) + P(H|S)P(S) = P(H|C)2P(F)+ P(H |F)P(F) + P(H|S)(1 – 3P(F)) = P(F)[2/4 + 2/5 –3/6] which gives P(F) = 5/24, P(C) = 10/24 = 5/12, and P(S) = 9/24 = 3/8. (b) The likelihood of a hit is P(H|F) = 2/5 or P(H|C) = 1/4, or P(H|S) = 1/6 depending on which hypothesis is true. Since P(H|F) = 2/5 is the largest, the maximum-likelihood decision is that it was a fastball. (c) Now we compare the joint probabilities P(H |F)P(F) = (2/5)(5/24) = 2/24, P(H|C)P(C) = (1/4)(10/24) = 2.5/24, and P(H|S)P(S) = (1/6)(9/24) = 1.5/24 and get the Bayesian decision that it was a curveball. Note that it is not necessary to find P(F|H), P(C |H), and P(S|H) explicitly; the joint probabilities suffice.