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FSRM 582, Homework 2 - Solutions Assigned: September 9, 2014 Due: September 16, 2014 Problem 1. (i) A factory runs three shifts. In a given day, 1% of the items produced by the first shift are defective, 2% of the second shift’s items are defective, and 5% of the third shift’s items are defective. If the shifts all have the same productivity, what percentage of the items produced in a day are defective? If an item is defective, what is the probability that it was produced by the third shift? (ii) If B is an event, with P (B) > 0, show that the set function Q(A) = P (A|B) satisfies the axioms for a probability measure. Thus, for example, P (A ∪ C|B) = P (A|B) + P (C|B) − P (A ∩ C|B) (iii) Show that if A and B are independent, then A and B c as well as Ac and B c are independent. (iv) Show that ∅ is independent of A for any A. (v) If A is independent of B and B is independent of C, then A is independent of C. Prove this statement or give a counterexample if it is false. (vi) If A and B are disjoint, can they be independent? (vii) If A ⊂ B, can A and B be independent? (viii) Show that if A, B and C are mutually independent, then A ∩ B and C are independent and A ∪ B and C are independent. (ix) A system has n independent units, each of which fails with probability p. The system fails only if k or more of the units fail. What is the probability that the system fails? (x) A player throws darts at a target. On each trial, independently of the other trials, he hits the bull’s-eye with probability .05. How many times should he throw so that his probability of hitting the bull’s-eye at least once is .5? 1 (xi) This problem introduces some aspects of a simple genetic model. Assume that genes in an organism occur in pairs and that each member of the pair can be either of the types a or A. The possible genotypes of an organism are then AA, Aa, and aa(Aa and aA are equivalent). When two organisms mate, each independently contributes one of its two genes; either one of the pair is transmitted with probability .5. a. Suppose that the genotypes of the parents are AA and Aa. Find the possible genotypes of their offspring and the corresponding probabilities. b. Suppose that the probabilities of the genotypes AA, Aa and aa are p, 2q and r, respectively, in the first generation. Find the probabilities in the second and third generations, and show that these are the same. This result is called the Hardy-Weinberg Law. Solution (i) a. Shifts have the same productivity =⇒ %(daily output defective) 1 of output from each of 3 shifts. 3 = P (defective|shift 1)P (shift 1) + P (defective|shift 2)P (shift 2) +P (defective|shift 3)P (shift 3) 1 · (0.01 + 0.02 + 0.05) = 3 8 . = 300 b. P (shift 3|defective) P (defective|shift 3)P (shift 3) P (defective) 1 0.05 · 3 = 5. 1 8 0.08 · 3 = = (ii) B ⊂ Q s.t. P (B) > 0, then Q(A) = P (A|B) is a probability measure. Proof: Must show that Q satisfies P M 1), P M 2) and P M 3). P M 1) Q(Ω) = 1: P (B) P (Ω ∩ B) By definition, Q(Ω) = P (Ω|B) = = = 1(since P (B) > 0). P (B) P (B) P M 2) Q(A) ≥ 0, all A ⊂ Ω: By assumption, P (B) > 0, and P is a probability measure, =⇒ P (A ∩ B) ≥ 0, all A ⊂ Ω. P (A ∩ B) Hence, Q(A) = ≥ 0. P (B) ∞ ∞ S P P M 3) A1 , A2 , . . . are disjoint, then Q( Ai ) = Q(Ai ). i=1 i=1 P (B ∩ (∪Ai )) P (∪i (Ai ∩ B)) P P (Ai ∩ B) P Q(∪Ai ) = = = i = i Q(Ai ). P (B) P (B) P (B) 2 P (A ∩ B c ) = P (A − B) |= B ⇒ A B c and Ac |= A |= (iii) B c Proof: = P (A − (A ∩ B)) = P (A) − P (A ∩ B) (byA ∩ B ⊂ A) = P (A) − P (A)P (B) (by independence) = P (A) [1 − P (B)] = P (A)P (B c ). |= Ac B c follows immediately [since (A ∪ B)c = Ac ∩ B c ]. (iv) P (∅ ∩ A) = P (∅) = 0 = P (∅)P (A), all A. (v) No. Take A = C in the statement of the problem. (vi) Yes if P (A) = 0 or P (B) = 0. (vii) Yes. Take B = Ω (the whole sample space). (viii) A, B, C are mutually independent. P [(A ∩ B) ∩ C] = P (A ∩ B ∩ C) = P (A)P (B)P (C) = P (A ∩ B)P (C). P [(A ∪ B) ∩ C] = P [(A ∩ C) ∪ (B ∩ C)] = P (A ∩ C) + P (B ∩ C) − P (A ∩ B ∩ C) = P (C)[P (A) + P (B) − P (A ∩ B)] = P (C)P (A ∪ B). (ix) P (failure) = = n X P (exactlyj units fail) j=k n X j=k n j p (1 − p)n−j j (x) P ( miss bull’s-eyes in each of n throws ) = 0.95n =⇒ P ( hit bull’s-eyes at least once in n throws ) = 1 − 0.95n . 1 log 2 So he should throw at least n∗ = − times to have probability to hit bull’s eye at least log 0.95 2 once. Since n is an integer, he needs at least 14 throws. 3 (xi) a. Either one of the pair is transmitted with probability 0.5. Possible genotypes AA Aa Probability 1 2 1 2 b. To find second generation probabilities given first generation probabilities p, 2q, r, for AA, Aa, aa. Can write table of conditional probabilities given parents’ genotype for second generation: Conditional P2 (AA) AA AA aa 1 1 2 0 Conditional P2 (Aa) AA AA 0 1 2 1 Aa Aa aa 4 Aa 1 2 1 4 0 aa Aa 1 2 1 2 1 4 aa 0 0 0 1 1 2 0 Conditional P2 (aa) AA AA 0 Aa 0 aa 0 Aa 0 1 4 1 2 aa 0 1 2 1 Unconditional probabilities:(law of cases) P2 (AA) = P2 (AA|AA, AA)P1 (AA, AA) + P2 (AA|AA, Aa)P1 (AA, Aa) +P2 (AA|AA, aa)P1 (AA, aa) + P2 (AA|Aa, AA)P1 (Aa, AA) +P2 (AA|Aa, Aa)P1 (Aa, Aa) + P2 (AA|Aa, aa)P1 (Aa, aa) +P2 (AA|aa, AA)P1 (aa, AA) + P2 (AA|aa, Aa)P1 (aa, Aa) +P2 (AA|aa, aa)P1 (aa, aa) 1 1 1 = 1 · p2 + · 2pq + 0 · pr + · 2pq + (2q)2 + 0 · 2qr + 0 · pr + 0 · 2qr + 0 · r2 2 2 4 = (p + q)2 . Similarly, P2 (Aa) = 2(p + q)(r + q) P3 (aa) = (r + q)2 . For the third generation, we see the same conditional probabilities as before but with genotype frequencies P2 (AA), P2 (Aa), P2 (aa) computed above. P3 (AA) P3 (Aa) P3 (aa) = (p + q)4 + 2(p + q)3 (r + q) + (p + q)2 (r + q)2 = (p + q)2 [(p + q)2 + 2(p + q)(r + q) + (r + q)2 ] = (p + q)2 . = 2[(p + q)2 + (p + q)(r + q) · (p + q)(r + q) + (r + q)2 ] = 2(p + q)(r + q)[(p + q + r + q) · (p + q + r + q)] (p + q + r + q = 1) = 2(p + q)(r + q). = 1 − P3 (AA) − P3 (Aa) = (r + q)2 . ♣ 5 Problem 2. [Powerball] In the PowerBall lottery game, 5 white, numbered balls are drawn uniformly at random, without replacement, from a set of 59 white balls numbered 1 to 59, and 1 red, numbered ball is drawn uniformly at random from 39 red balls numbered 1 to 39. To play the game you buy a $1 ticket with 5 distinct numbers from 1 to 59 (to match the white balls) and 1 special number (the “powerball”) from 1 to 39 (to match the red ball). If you match exactly 3 of the white balls plus the red ball, you win fifth prize, worth $100. If you match exactly 4 of the white balls and not the red ball, you win fourth prize, also worth $100. 1. What is the probability to win fifth prize? 2. What is the probability to win fourth prize? 3. For this part, suppose you buy two tickets for the same drawing. (This will cost $2.) Suppose that on the first ticket, you pick 1, 2, 3, 4, 5 as your white numbers and 6 as your red number, while on your second ticket, you pick 51, 52, 53, 54, 55 for white and 6 again for red. Find the chances that you win: a fifth prize on both tickets; a fifth prize on exactly one ticket; a fifth prize on at least one ticket. Solution 1. To win 5th prize, you must match exactly 3 of 5 white balls plus the powerball. Let E5 be the event you win the 5th prize on a single ticket. 5 59 − 5 1 14310 1 2 1 3 = ≈ . P (E5 ) = 59 39 195, 249, 054 13644 1 5 2. To win the 4th prize, you must match exactly 4 of white balls but not the powerball. Let E4 be the event you win the 4th prize on a single ticket. 5 59 − 5 39 − 1 10260 1 4 1 1 = ≈ . P (E4 ) = 59 39 195, 249, 054 19030 5 1 3. Let E51 be the event you win the 5th prize on ticket {1, 2, 3, 4, 5}{6}, and E52 be the event you win the 5th prize on ticket {51, 52, 53, 54, 55}{6}. Winning 5th on both tickets require matching exactly 3 numbers on each ticket and is the event E51 ∩ E52 = ∅, so P (E51 ∩ E52 ) = 0. Since E51 ∩ E52 = ∅, we have P (E51 ∪ E52 ) = P (E51 ) + P (E52 ) = 2P (E5 ). P (5th on at least one ticket) = P [E51 ∩ (E52 )c ] + P [(E51 )c ∩ E52 ] + P (E51 ∩ E52 ) = P (E51 ) + P (E52 ) = 2P (E5 ) = P (5th on exactly one ticket). 6 ♣ Problem 3. [Random Coins] Suppose you have three coins, which we will call Coin I, Coin II and Coin III, each with a different probability of coming up heads on a standard flip of the coin. Coin I comes up heads 1/3 of the time, Coin II comes up heads 1/2 of the time and Coin III comes up heads 3/4 of the time. Below, you are given different situations involving a scheme for flipping two coins and recording the outcome (Heads or Tails) of each flip in order. For each scheme, answer the following series of questions. (0) What is the sample space of this experiment? (i) Are the two flips independent? Explain. (ii) What is the probability that one head and one tail (in either order) is flipped? (iii) What is the probability of heads on the second flip? (iv) What is the probability that a head was flipped on the first flip given that a tail was flipped on the second flip? Answer (i)-(iv) for each of the following scenarios (1)-(3). Write {I, II, III} to denote the set of Coin I, Coin II and Coin III. 1. Choose a coin uniformly from {I, II, III}, flip it, record the outcome, replace the coin and repeat once. 2. Choose a coin uniformly from {I, II, III}, flip it, record the outcome, do not replace the coin; for the second flip, choose a coin uniformly among the remaining (unchosen coins), flip it, record the outcome. 3. Choose a coin uniformly from {I, II, III} and flip it twice, recording the outcome. (v) Explain why scenarios 1 and 3 differ. Be as concise and precise as possible. Solution (0) The sample space is Ω = {HH, HT, T H, T T }. Note that we do not which coins are used, so that is not part of Ω. 7 Let’s look at (i)-(iv) for each of the 3 scenarios (1)-(3) separately. (1) (i) Yes, the flips are independent. (ii) Overall probability of heads on single flip, PH , is 1 1 1 1 3 19 PH = · [P (H| coin I) + P (H| coin II) + P (H| coin III)] = [ + + ] = . 3 3 3 2 4 36 19 · 17 = 0.498. Hence, P {(H, T )} = 2 362 19 (iii) P ( H on 2nd) = P ( H on 1st) = . 36 19 . (iv) P ( H on 1st| T on 2nd)) = P ( H on 1st) ( by independence ) = 36 (2) (i) No, coins are now chosen without replacement so the choice of the first coin affects the outcome of subsequent flips. (ii) We can compute P ({H, T }) by conditioning on the choice of the first coin. For each coin, P (H|coin) is given above. Now, we consider P (H on 2nd flip | coin y on 1st flip), for y = I, II, III P (H, 2nd|I) = P (H, 2nd|II) = P (H, 2nd|III) = 1 1 3 5 [ + ]= 2 2 4 8 1 1 3 13 [ + ]= 2 3 4 24 1 1 1 5 [ + ]= 2 3 2 12 Hence, P (H, T ) = = = 1 · [P (H, 1st|I)P (T, 2nd|I) + P (H, 1st|II)P (T, 2nd|II) + P (H, 1st|III)P (T, 2nd|III)] 3 13 5 1 3 5 1 1 · [ · (1 − ) + · (1 − ) + · (1 − )] 3 3 8 2 24 4 12 38 . 144 By exchangeability, P ({H, T }) = 2P (H, T ) = (iii) P (H on 2nd flip ) = P (H on 1st flip ) = 8 19 . 36 76 . 144 (iv) P (H on 1st|T on 2nd) (3) = P (H on 2nd|T on 1st) P (T, H) = P (T ) 38 = 144 17 36 38 = . 68 (i) No, flips are not independent. They are conditionally independent given the choice of coin. (ii) 1 · 3 1 P ({H, T }| coin II) = 2 · · 2 3 P ({H, T }| coin III) = 2 · · 4 1 4 ·( P ({H, T }) = 3 9 P ({H, T }| coin I) = 2· 2 4 = 3 9 1 1 = 2 2 1 6 = 4 16 1 6 95 + + )= . 2 16 216 19 ; 36 (iv) Given the outcome of the second flip, we have a different conditional probability that the coin that we are flipping is coin I, II, or III. For example, if you tell me the 2nd 2 flip is a tail, I’m more inclined to think you are flipping coin I (with probability of 3 1 tail) than coin III (with probability of tail). 4 (iii) P (H on 2nd) = PH = P ( coin I|T on 2nd) = = = 9 P (T on 2nd|I)P ( coin I) P (T on 2nd) 2 1 · 3 3 19 1− 36 8 . 17 P ( coin II|T on 2nd) = = = P ( coin III|T on 2nd) = = = P (T on 2nd|II)P ( coin II) P (T on 2nd) 1 1 · 2 3 19 1− 36 6 . 17 P (T on 2nd|III)P ( coin III) P (T on 2nd) 1 1 · 4 3 19 1− 36 3 . 17 Hence, P (H on 1st|T on 2nd) = P (H on 1st|I, T on 2nd)P (I|T on 2nd) +P (H on 1st|II, T on 2nd)P (II|T on 2nd) +P (H on 1st|III, T on 2nd)P (III|T on 2nd) 1 8 1 6 3 3 = · + · + · 3 17 2 17 4 17 95 = . 204 ♣ 10