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Unformatted text preview: CHAPTER 2 2—1 We use De Morgan's law:
(a) K+§+ 3+3 = AB + A3 = A(B+'B) = A
(b) (A+B)(KF) = (A+B)(A‘+§) = A§+ BK
because A; = m m? = w} ' WW”...
22 IfA={2f_x_<_5} B={3fx:6} S={°°<x<°°} then
A+B={2§x§6} AB={3§x§5}
(“mm = {29:56} [{x<3} + {x>5}]
={2§x<3}+{5<x_<_6} 23 If AB = {m} then ACE hence P(A) 5w?)
2—4 (a) P(A) = P(AB) + MAE) pas) = P(AB) + P(KB)
If, therefore, P(A)==P(B)  P(AB) then
P(Aﬁ) = 0 P(KB) = 0 hence P(EBHE) = P(KB) + P(Ai)  o
(b) If P(A) = P(B) = 1 then 1 = P(A)§ P(AFB) hence
I = P(A+B) = P(A) + P(B)  P(AB) = 2  P(AB)
This yields P(AB) = l 25 From (2—13) it follows that
P(A+B+C) = P(A) + P(B+C)  P[A(B+C)]
P(B+C) = P(B) + P(C)  P(BC)
P[A(B+C)] = P(AB) + P(AC)  P(ABC)
because ABAC = ABC. Combining, we obtain the desired result. Using induction, we can show similarly that P(A1+A2+ +An) = P(Al) + P(A2)+ ...+P(An)  P(AlAz) — _ P(An_1An) + P(A ) + ...+P(An2An) 1A2A3 *_*1’(A1A2  ‘  An) w. ...... w— m... V... n... .W 26 Any subset of S contains a countable number of elements, hence, it
can be written as a countable union of elementary events. It is therefore an event. 27 Forming all unions, intersections, and complements of the sets {1} and {2,3}, we obtain the following sets:
{6}, {1}, {4}, {2,3}, {1,4}, {1,2,3}, {2,3,4}, {1,2,3,4} 28 If ACB,P(A) = 1/4, and P(B)  1/3, then
P(AB) a pm _, 1/4 : 3 P(AIB)=_P(B) Pm m z
_ P(AB) a P(A) _
P(BA) _ P(A) P(A) ' 1
29 P(ABC)P(BC) , %£%&%1 zﬂﬁgg
P(ABC)
a P(C) = P(MIC) P(ABC) P(BC)
P(BC) P(C) P(ABC)P(BC)P(C) = P(C) = P(ABC) 210 We use induction. The formula is true for n==2 because P(A1A2) = P(A2IA1)P(A1). Suppose that it is true for n. Since p(A ...A1) = P(A ...A2A1)p(A1 ...An) n+lAn n+lAn we conclude that it must be true forr1+1. 211 First solution. The total number of m element subsets equals (3) (see Probl. 226). The total number of m element subsets containing :0 equals n—l
(m_l). Hence n nl m
p a(m)/(rnl) = E Second solution. Clearly, P{coIAm} = m/n is the probability that to
is in a specific Am. Hence (total probability) m m
p = X PicolAm}p(Am) = a Z P(Am) = n where the summation is over all sets Am. v“ 2—12 (a) P{6ft§8}=1% (b) P{6:t§8t>5} , 2136:1258} 3% 213 From (227) it follows that to+t1
P{toft§to+tlltzto} 'J aunt/I a(t)dt
t t o 0
t1
PM: 5 t1}= I a(t)dt
O
Equating the two sides and setting tlto+ At we obtain a(to)/J a(t)dt = «1(0) t
o
for every to. Hence,
” ” «(0):o
 in J a(t)dt = a(0)to I a(t)dt  e
t t
o o Differentiating the setting c=a(0), we chclude that
ct Ct P{t§t1}=1e 1 a(to) = c e W 214 If A and B are independent, then P(AB) =P(A)P(B). If they are
mutually exclusive, then P(AB)  0. Hence, A and B are mutually
exclusive and independent iff P(A)P(B)  0. WWW 1 = AlAZ + AlA2 hence P(A1) = P(A1A2) + P(A1A2) 215 Clearly, A If the events Al and K2 are independent, then P(A1A2) = P(A1) — P(A1A2) = P(A1)  P(A1)P(A2)
 P(A1)[1 —P(A2)] = P<A1)P(K2) hence, the events A1 and K2 are independent. Furthermore, S is independent with any A because SA  A. This Yields P(SA) = NA) = P(S)P(A)
Hence, the theorem is true for n=2. To prove it in general we use induction: Suppose that An+1 is independent of A An. Clearly, 1,000,
An+1 and An+1 are independent of Bl....,Bn. Therefore P(Bl N . BnAn+l) I P(Bl ' O . Bn)P(An+l) 1:031 ... BnAn+1) = P(Bl 'H Bn)P(An+1) 2.16 The desired probabilities are given by m ~ 1
k: — 1
n
k 2.17 Let 141,142 and A3 represent the events A1 2 “ball numbered less than 07" equal to m is drawn”
A2 : “ball numbered m is drawn”
A3 :: “ball numbered greater than m is drawn” P(A1 occurs n1 2 k — 17 A2 occurs n2 2 1 and A3 occurs n3 2 0)
(m + 712 + _ ml 1712 ng
t“ m" " ‘ P p I):
n1! n2! n3! 1 2 5 I fh)! (Swill—1 (k <33)“  l 2.18 All cars are equally likely so that the ﬁrst car is selected with
probability p : 1/3. This gives the desired probability to be (if) (if (if = 0’26 2.19 P{“d7‘awmg a white ball ’7} 2 P(“atleat one white ball m k trials ” n
)
z 1 — P(“ull black balls in k trials”) (“1 : 1....j...
m+n
( k > 2.20 Let D = 27’ represent the penny diameter. 80 long as the center
of the penny is at a distance of 7“ away from any side of the square,
the penny will be entirely inside the square. This gives the desired
probability to be @3341“?le 2.21 Refer to Example 3.14.
(a) Using (3.39), we get (b) P(“tw0 one—digit and four two—digit numbers”) 2 P(“all one — digit numbers”) 2 . n
The number of equatlons of the form P(AiAk) P(A1)P(Ak) equals (2). The number of equations involving r sets equals (g). Hence the total number N of such equations equals
n n n
. = +...
v <2)+(3) +(n) And since n n n _ :1: n
(0)+(1)++(n)  (1+1) 2 we conclude that n=2“ (3)<‘1‘)=2“1n 1 and B2 by R the set of red balls. We denote by B respectively the balls in boxes 1 and 2 and We have (assumption) P(Bl)=P(B2)=0.5 P(RIBl)=O.999 P(R32)=o.001 Hence (Bayes' theorem) P(RI31)P(B1) 0.999 “31‘” " P(R31')‘p"(n'1‘)+“'p'(“]n nzmnz) g o.999+o.oo1 = 0'9” WWW 6 2—24 We denote by Bl and 32 respectively the ball in boxes 1 and 2 and by D all pairs of defective parts. We have (assumption)
P(B1) = P(BZ) = 0.5 To find P(D[Bl) we proceed as in Example 2—10: First solution. In box B1 there are 1000 x999 pairs. The number of pairs with both elements defective equals 100 x 99. Hence, 100 X 99
1000 X 999 Second solution. The probability that the first bulb selected from Bl is defective equals 100/1000. The probability that the second is defective assuming the first was effective equals 99/999. Hence, P(DBl) = 100 99
P(Dl31) 1000 x 999
We similarly find
_ 100 99
P(DBz) ' 2000"1999 (a) P(D) = P(DBl)P(Bl) + P(DI32)P(BZ) = 0.0062 P(DBl)P(Bl)
9(0)
_____n_______________,_______________________________________________________ (b) P(BllD) = = 0.80 2—25 Reasoning as in Example 213, we conclude that the probability that the bus and the train meet equals 2 2
10 x
(10+x)60—2 ——2 Equating with 0.5, we find x= 60 lO/ll. 2—26 We wish to show that the number Nn(k) of the element Subsets of S equals
a n(n—l) "' (nk+l)
Nn(k) "ITTfTTT"T?" This is true for kl because the number of lelement subsets equals n. Using induction in k, we shall show that nk
k+1 We attach to each kelement subset of S one of the remaining nk elements Nn(k+1) = Nn(k) 1<k<n (1)
of S. We, then, form Nn(k) (nk) k+1element subsets. However, these subsets are not all different. They form groups each of which has k+l
identical elements. We must, therefore, divide by k+1. '7 2—27 In this experiment we have 8 outcomes. Each outcome is a selection of a particular coin
and a specific sequence of heads or tails; for example fhh is the outcome "weselected the
fair coin and we observed hh". The event F = {the selected coin is fair} consists of the
four outcomes fhh, fht, fth and fhh. Its complement F is the selection of the two— headead coin. The event HH = {heads at both tosses} consists of two outcomes. Clearly, P(F) = P(_) = —;— P(HHF) = 7:— P(HHF) = 1 Our problem is to find P(FHH). From (241) and (243) it follows that P(HH) = P(HHF)P(F) + P(HHI:)P(F) = —3
P(HHF)P(F) _ 1/4 x 1/2 _ l
P(F'HH)=—Hﬁﬁ3— ‘—‘s/T "—5" ...
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Harris

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