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hw1sol - Introduction to Probability and Statistics I Stat...

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Unformatted text preview: Introduction to Probability and Statistics I Stat 42'? Homework 1 Solution 1.1.2 3 : {0 females, 1 female, 2 females, 3 females, . . ., 11 females} 1.1.5 5 2 {[red, shiny}, [13:1, dull]. (blue. shiny), [blue, dullj} 1.1.111 PH) = 2 :4 P111) and P111) = 3 X Pan} => P111 = 6 >1 mm). Then 9(1) + P(I]] + P|[I[1} : 1 :> {a 3-: 131111)] + {s 1-: P(Il]]]| + P[[II} = 1. Hence P[I]]) = rig, P111} = 3 >1 P1111}. = g, and P[I} = a x mm}. = ,2, 1.2.2 (a) PTA) 2 P131} + Pic) + Pfle) : 0.27 so Pit) +£1.11 + [1.116 = 0.27 and hence Pfit} = [1.11:1 [b] 131.1“) = 1 — PEA) = 1 — 0.27 = 11.73. [c] P[A'} = Phil} +P{d) + P[f] = U33 so 0.09 +P[d] + [1.29 = llTS and hence 1?de = 11.35. 1.2.6 In Figure 1.10, let (1', 11] represent the outcome that the score on the red die is :r and the score on the blue die is y. The event that the score on the red die is strictly greater than the score on the blue die consists of the following 15 outcomes: {(211}! (3:1): (312]! [471}.- (412L (4:3)! (5:1): [1512]: {5:3}: (514]: (611)! [1652)! [6:3]: {514). (5,5)} The probability of each outcome is 3%, so the required probability is 15 X 3—16 2 1—52. This probability is less than 0.5 because of the possibility that both scores are equal. The complement of this event is the event that the red die has a score less than. or equal to the score on the blue die with a probability of 1 — 15—2 : %. 1.2.10 1.3.2 1.3.4 [0.) See Figure 1.24. P{'Typo I batter}r lasta longest} = PHI], l1]. 1)} -|— PI[I[111,11, [)1 = 0.30 + 0.03 = 0.42. {b} P('1‘ypo I batter}r laste shortest] = PHI, [1. III” -|— PHI, III, [1)] = 0.11 + 0.0? = 0.10. [e] Plnype I bath—1r}r does: not last longest] : 1 — P['I‘3rpe 1 battery laete longest] = 1 — 0.42 = 0.58. (d) PfI‘ype I |:I»a1‘.’oer_r}r last longer than Type 11] = P([11. 1. III” + P[[11.111.1)) + Pii111,11.1]] = 0.24 —l— 0.30 + 0.03 = 0.00. {a} See Figure 1.55. P{B} = 0.01 + 0.02 + 0.05 + 0.11 + 0.03 + 0-00 + 0.13 = 0.40. {h} HR 0 C] = 0.02 + 0.00 + 0.11 = 0-10. (0) P{A LI C] = 0.0? —|— 0.05 —|— 0.01+ 0.02 —|— 0.05 —|— 0.08 + 0.04 + 0.11 + 0.07 + 0.11 = 0.01. [0} P01 0 B o C) = 0.02 + 0.00 = 0.07. (e) P(A u B u C) = 1 — 0.00 — 0.04 — 0.05 = 0.00. {f} P{A’ m a) = 0.00 + 0.00 + 0.11 + 0.10 = 0.00. (g) P(B’ U C} = 0.04 + 0.03 + 0.05 + 0.11 + 0.05 + 0.02 + 0.03 + 0.04 + 0.11 + 0.07 + 0.07 + 0.05 = 0.72. ([1) P01. 0 (0 r1 (3)) = 0.07 + 0.00 +0.01 + 0.02 + 0.00 + 0.00 + 0.04 + 0.11 = 0.43. 0) ma 0 B) n o} = 0.11 + 0.00 + 0.02 + 0.00 + 0.04 = 0.30. (j) PfiA'UCP} = 1 — P{A’ L1 0} Phi" U0) = 0.04 + 0.03 + 0.05 + 0.00 + 0.00 + 0.13 + 0.11 + 0.11+ 0.07 + 0.02 + 0.05 + 0.03 + 0.04 = 0.37. 136.410 0)“) = 1 _ 0.07 = 0.13. A U C’ : {all females together 1with an)r man who does not ham brown eyes}. A“ PI 13 ["I C} 2 {males with black hair and brown eyes}. A H [13 U C) = [1611101105 with either black hair or brcmrn eyes or but-11}. Noie: Stodenis can tryr to use suggesiiue notations in formulating your own problems. F0;r exampie, F=[personi00female]. elc. 1.3.6 1.3.12 _ 1.4.4 1.4.6 1.4.8 P{AuB)= P(A)+P(B)— -—P(AnB_) < 1 so P(B)_<_1—04+3.3 : 0.9.. 3133 11(11):» P(AnB)— _ 0.3 50 9.3 < 13(3) < 39. ‘ Let the event. R be that'a red ball is chosen and let'the event 8 be that a shiny ball is chosen. 1 It is known that P(Rn 31:55, P(S)= fiend P(R)= Hence the probability that the chosen ball'm is either shiny or red is P(RUS)=P(R)+P(S)—P(Rn3)= 5951-5203- = E5';=3.373. The probability of a dull blue b81113 P(R’ n S') = 1 - P(R U S) = 1— 0. 575— — 0. 425. PM) is 311131191111311 PM | 11). Event 3 13 a. necessary condition for event A and 90' cmditioning on event B hicreesea the 13101131111113 of event A. LettheeventObeanontamerepmrandlettheeventsbeasatmfactorvrepmr ItiskmwnthatP(S|O}= 0.85311dP(O): (1“??? The question 35133 for P(On S) .131111111'115 130:) n 3) = pm | 0) 313(9) = 0.33 x 0.77 = 11.3343. 01.193311111339931 petiod1neludtngone 1911113931: the nu1nber 016331313 {3 33651+366 31,461. 01’ these 4 x 1.2 = 48 occur on the first day of ‘3. month and so the probability that a birthday falls on the first day of a month is W 2: 0.0929. 01‘ theee 1,461 days 4 K 31 = 124 009111 in M91911 of which 4 days are Math lst. Consequently. the probability that a. birthday falls on March 131:. conditional that it 13111313mh131§=gr=39333 Ofthese 1.461(1335 (3x 28) + 29 = 113 occur in February of which 4 days are. ‘ February 131;. Cbnsequaifly, the probability that a birthday falls on February 131;. conditional 3113111 is in February is I}; = 11.9354. ...
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