mont4e_sm_ch02_supplemental - P( A | B) = 2-121....

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Supplemental Exercises 2-125. Let B denote the event that a glass breaks. Let L denote the event that large packaging is used. P(B)= P(B|L)P(L) + P(B|L')P(L') = 0.01(0.60) + 0.02(0.40) = 0.014 2-126. Let "d" denote a defective calculator and let "a" denote an acceptable calculator a a) {} aaa daa aad dad ada dda add ddd S , , , , , , , = b) daa dad dda ddd A , , , = c) ada add dda ddd B , , , = d) dda ddd B A , = e) aad dad ada add dda ddd C B , , , , , = 2-21
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2-127. Let A = excellent surface finish; B = excellent length a) P(A) = 82/100 = 0.82 b) P(B) = 90/100 = 0.90 c) P(A') = 1 – 0.82 = 0.18 d) P(A B) = 80/100 = 0.80 e) P(A B) = 0.92 f) P(A’ B) = 0.98 2-128. a) (207+350+357-201-204-345+200)/370 = 0.9838 b) 366/370 = 0.989 c) (200+163)/370 = 363/370 = 0.981 d) (201+163)/370 = 364/370 = 0.984 2-129. If A,B,C are mutually exclusive, then P( ) = P(A) + P(B) + P(C) = 0.3 + 0.4 + 0.5 = ABC ∪∪ 1.2, which greater than 1. Therefore, P(A), P(B),and P(C) cannot equal the given values. 2-130. a) 345/357 b) 5/13 2-131 (a) P(the first one selected is not ionized)=20/100=0.2 (b) P(the second is not ionized given the first one was ionized) =20/99=0.202 (c) P(both are ionized) =P(the first one selected is ionized) × P(the second is ionized given the first one was ionized) =(80/100) × (79/99)=0.638 (d) If samples selected were replaced prior to the next selection, P(the second is not ionized given the first one was ionized) =20/100=0.2. The event of the first selection and the event of the second selection are independent. 2-132. a) P(A) = 15/40 b) P( BA ) = 14/39 c) P( A ) = P(A) P(B/A) = (15/40) (14/39) = 0.135 B d) P( ) = 1 – P(A’ and B’) = AB 25 24 1 0.615 40 39 ⎛⎞ −= ⎜⎟ ⎝⎠ A = first is local, B = second is local, C = third is local e) P(A B C) = (15/40)(14/39)(13/38) = 0.046 f) P(A B C’) = (15/40)(14/39)(25/39) = 0.089 2-133. a) P(A) = 0.03 b) P(A') = 0.97 c) P(B|A) = 0.40 d) P(B|A') = 0.05 e) P( ) = P( )P(A) = (0.40)(0.03) = 0.012 f) P( ') = P( ' )P(A) = (0.60)(0.03) = 0.018 g) P(B) = P( )P(A) + P( ')P(A') = (0.40)(0.03) + (0.05)(0.97) = 0.0605 2-134. Let U denote the event that the user has improperly followed installation instructions. 2-22
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Let C denote the event that the incoming call is a complaint. Let P denote the event that the incoming call is a request to purchase more products. Let R denote the event that the incoming call is a request for information. a) P(U|C)P(C) = (0.75)(0.03) = 0.0225 b) P(P|R)P(R) = (0.50)(0.25) = 0.125 2-135. (a) 18143 . 0 ) 002 . 0 1 ( 1 100 = = P (b) 005976 . 0 002 . 0 ) 998 . 0 ( 2 1 3 = = C P (c) 86494 . 0 ] ) 002 . 0 1 [( 1 10 100 = = P 2-136. P( ) = 80/100, P(A) = 82/100, P(B) = 90/100. AB Then, P( A ) P(A)P(B), so A and B are B not independent. 2-137. Let A i denote the event that the ith readback is successful. By independence, PA A A PA PA PA () ( ) ( ) ( ) ( . ) . '' ' ' ' ' 123 1 2 3 3 0 02 0 000008 ∩∩ = = = .
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This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.

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mont4e_sm_ch02_supplemental - P( A | B) = 2-121....

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