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Unformatted text preview: Introduction to Probability and Statistics I Stat 411' Homework 1 Solution 1.1.2 8 = {ID fenntles. 1 female, 2 ietnules, 3 females. 'H feniules} 1.1.13 5 = {[I'EELL shiny}, [L'eLL dull). {blue shiny]. [blue dullj} LMU HD=2xPMlmﬂPﬂh=3xPMH : HD=ﬁXPMH "l‘hen em + em] + l‘t_]lj :1 :> to x mu}; + .33 >< emu] + emu. = 1.
Hence Pull] = ﬁ. Pin] = s x PfIII'i = 1—5;] and em = :3 X Pan; = 11”.
12 2 [:11] PEA] : Pity] + Ptr'j + P[r—:] 211.27 H“ PIli] +1111 + illIii : 11.27 .51.tl lierice
PHI} : 0.11}.
[h] NIL/1'] = 1 — l’Ex'l] =1 — [1.27 = [1.73.
[cl FLA”) = + PM) + PU] = LI.?3 so [1.09 + Pitt] + LL29 = ULTS and hence
PM] : 1135. 1.2.6 I11 Figure 1.10. let (I. in represent the outcome that the score on the reel die is ;r
and the score on the hlue rlie is _i.r. The event that. the secure on the rerl die is ei'r‘irtlgr
greater than the score on the blue die consists of the following 15 outcomes: {(2.1}, (3.1). [3.2], [4.1). (4.2}, [4.3). (5.1), [5.23, (5.3}, [5,4], [6,1]. (6,2), [6,3].
(54), [5,5]}
The probability of each outcome is so the required probability is 15 >< % = This probability is less than 43.5 because of the possibility that both scores are equal.
The complement of this event is the event that the red the has 3. score less than or 4..
I equal to the score on the blue die with a probability: of 1 — : E. 152.10 1:11} Soo Figure 1.24.
P[T§,13o I butt1'3; lasts ]ongost] = P1111, 11]. [1:1 + 1.311111, 1], [11:1
= 0.30 + 0.03 = 0.42.
1:13} PCT}qu I batter}; lasts shortest] = PI: [1. 11. [11]] + PHI. 111. [1]]
= 0.11 + 0.0? = 0.18. (o) Pfl‘y‘pe 1 battery does not last longest] : 1 — PUPVpe 1 battery; Lasts longest)
= 1 — 0.42 = 0.08. 11:1} 1301531312 1 better}; last longer than r11312113. 11}
= P1111. i. 111]] + P[[11. 111. 1]] + P11111. 11. = 0.31 + 0.30 + 0.03 = 0.00. 1.3.2 1:11] Soo Figure 1.5.5.
131le = 0.01 + 0.02 + 0.05 + 0.11 + 0.03 + 000 + 0.13 = 0.40. [11} P113 1*: C] = 11.112 + 11.115 + 11.11 = 11.111. ['61 P{A LI [‘1 = 11.11? + 11.115 + 11.111 + 11.112 + 11.115 + 11.1111 +11.114 + 11.1] + 11.11? +
11.11 = 11.111. [1]} P19. 1'1 B 11 C} = 11.112 + [1.1111 = 11.117.
(12‘; HA U 13 11 C1) = 1 — 11.113 — [1.114 — 11.115 =1].111>1.
(1'1 P{A’ m 13) = [1.1111 + 11.1111+ 11.11 + 11.111 2 11.1111. (14] P{B" 1.1 C] = 0.04 + 0.03 + 0.05 + 0.11 + 0.05 + 0.02 + 0.08 + 0.04 + 0.11 +
{1.1}? + 0.0? + 0.05 = 0.72. [11} P191. 1.] (13 ['1 C1] : (1.117 + 0.05 +0.01 + 0.02 + 0.05 + 0.01.1 + [HI1 + [1.11 :
0.43. [11 P11}. 1.1 B1 1" 131 = 11.11 + 0.115 + 11.02 + [108 + 11.0.1 = 11.311.
(j) P614“ 1111]“) = 1 — P{A’ 11 C} P111060: 0.01 + 0113+ 0.05 + 0.015 + 0.00 + 0.13 + 0.11 + 1111+ 0.07 +
0.02 + 0.05 — 0.08 + 0.01 2 0.87. 13611111 1:11“) 2 1 _ 11.117 2 11.13. 1.3.]. [a] A 1'1 13 = {females wiﬂl 1:11st hair}.
[17] .151. U C" : {11.11 females together with any 11111.11 who does not have brown eyes}.
[c] A“ ['1 B ['1 C‘ : {males with black hair and brown eyes}. (1.1] 151. D [B U C) : {161.115.1165 with either black 1111.11 or 111011111 eyes 111' 111.1111}. Note: Siudents can try to use suggestive notations in formulating your own
problems. Forexemple, lepersonisafemaie], e111. 1.3.6 1.3.12 _ 1.4.4 1.4.6 1.4.8 pmue) = P(A)_+P(B)—P(Anﬁ)_g 1 50pm) _<_ i—o.4+e.3 0.9'
Also, PUB) 2; P(AnB) = 0.3. so 0.3 <_: Pa?) 5 0.9. ‘ Let the event. R be that'a red bell is chosen and let'the event 8 be that a shiny hall
is chosen. ' It is known that Pam S) = $5, 19(3) = %and P(R) = gga.
Hence the probability that the chosen ball is either shiny or red is P(RU.S') = P(R)+P(S)—P(ROS) = 5%+§%§% = g; = 0.575. The probability of a dull blue banfis P(R’ n '5') = 1 H P(R u s)
.= 1 — 0.575 = 0.425.  PM) is than PM  B). Event 3 is a. necessary condition for event A and so
cmditioning on event B increases the probability of event A. Let the Event 0 be anon time repair and let the event 8 be ‘a. satisfactory It is known that P(._S  O} = 0.85 and 13(0) = 0.77. The qnieetion cake for PIOI’! S) is 131:0 n 3) = pm we xP(Q) = 0.35 x 0.77 = 0.5545. Over afoul“ year pelted including one leap year; number of days is {3 x 365] + 366 21,461. _ __  Of these 4 x 1.2 = 48 occur on the ﬁrst dear of a. “month and so the prrolzmhilit1,r
that a birthday falls on the ﬁrst day of a month is _‘3 =: 0.0329. ' Of these 1,461 days 4 K 31 = 124 occux in March of which 4 days are March lst.
Consequently. the probability that a. birthday falls on March lst. conditional that it
ismMerchisIﬁ=§T=Q03n Ofthese 1.461 days (3x 28) + 29 = 113 occur in February of which 4 days are ‘
February let. Cbnsequenﬂy, the probability that a birthday falls on February let.
conditional that it is in February is I}; = 0.0354. ...
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 Spring '08
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 Probability

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