Quiz 1.1 solution In the Venn diagrams for parts (a)-(g) below, the shaded area to represent the indicated set. OMTT
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Quiz 1.1 solution In the Venn diagrams for parts (a)-(g) below, the shaded area to represent the suggested set. O M T T (a) N = T c (b) N ∪ M O M O M O M T T (d) T c ∩ M c (c) N ∩ M Quiz 1.2 solution A1 B1 A2 B2 A3 B3 = vvv, vvd, dvv, dvd = vdv, vdd, ddv, ddd = vvv, ddd = vdv, dvd = vvv, vvd, vdv, dvv, vdd, dvd, ddv = ddd, ddv, dvd, vdd Recall that Ai and Bi are collectively exhaustive if Ai ∪ Bi = S. Also, Ai and Bi room mutually exclude, if Ai ∩ Bi = φ. Because we have written down each pair Ai and Bi above, we can simply examine for this properties. The pair A1 and B1 are mutually exclusive and collectively exhaustive. The pair A2 and B2 are mutually exclusive however not jointly exhaustive. The pair A3 and B3 space not support exclusive since dvd belongs come A3 and B3 . However, A3 and also B3 are jointly exhaustive. 2 trouble 1.1.1 Solution based on the Venn diagram on the right, the finish Gerlandas pizza food selection is • regular without toppings • consistent with mushroom • regular with onions • constant with mushrooms and also onions • Tuscan without toppings • Tuscan with mushrooms O M T trouble 1.1.2 Solution based upon the Venn diagram on the right, the answers space mostly relatively straightforward. The just trickiness is that a pizza is one of two people Tuscan (T ) or Neapolitan (N ) for this reason N, T is a partition yet they space not depicted as a partition. Specifically, the event N is the an ar of the Venn diagram external of the “square block” of occasion T . If this is clear, the concerns are easy. O M T (a) since N = T c , N ∩ M 6= φ. Hence N and also M are not support exclusive. (b) Every pizza is either Neapolitan (N ), or Tuscan (T ). Thus N ∪ T = S so the N and also T are collectively exhaustive. Therefore its likewise (trivially) true that N ∪ T ∪ M = S. That is, R, T and M are also collectively exhaustive. (c) indigenous the Venn diagram, T and O space mutually exclusive. In words, this means that Tuscan pizzas never have actually onions or pizzas through onions are never Tuscan. Together an aside, “Tuscan” is a fake pizza designation; one do not do it conclude that human being from Tuscany in reality dislike onions. (d) indigenous the Venn diagram, M ∩ T and O space mutually exclusive. Thus Gerlanda’s doesn’t do Tuscan pizza with mushrooms and onions. 3 (e) Yes. In regards to the Venn diagram, these pizzas space in the collection (T ∪ M ∪ O)c . trouble 1.1.3 equipment R in ~ Ricardo’s, the pizza tardy is either roman (R) or Neapolitan (N ). To draw the Venn diagram on the right, we make the following observations: W N M O • The collection R, N is a partition so us can draw the Venn diagram with this partition. • only Roman pizzas can be white. Therefore W ⊂ R. • just a Neapolitan pizza can have onions. Therefore O ⊂ N . • Both Neapolitan and also Roman pizzas have the right to have mushrooms so that event M straddles the R, N partition. • The Neapolitan pizza can have both mushrooms and also onions therefore M ∩ O can not be empty. • The problem statement does not preclude placing mushrooms ~ above a white roman pizza. Hence the intersection W ∩M must not it is in empty. difficulty 1.2.1 equipment (a) an end result specifies even if it is the connection speed is high (h), tool (m), or short (l) speed, and also whether the signal is a computer mouse click (c) or a tweet (t). The sample an are is S = ht, hc, mt, mc, lt, lc . 4 (1) (e) The event that much more than one circuit is agree is C = aaa, aaf, af a, f aa . (4) The occasion that at the very least two circuits failure is D = f f a, f af, af f, f f f . (5) (f) Inspection mirrors that C ∩ D = φ for this reason C and D room mutually exclusive. (g) because C ∪ D = S, C and D are jointly exhaustive. difficulty 1.2.3 equipment The sample an are is S = A♣, . . . , K♣, A♦, . . . , K♦, A♥, . . . , K♥, A♠, . . . , K♠ . (1) The occasion H is the collection H = A♥, . . . , K♥ . (2) difficulty 1.2.4 systems The sample an are is 1/1 . . . 1/31, 2/1 . . . 2/29, 3/1 . . . 3/31, 4/1 . . . 4/30, 5/1 . . . 5/31, 6/1 . . . 6/30, 7/1 . . . 7/31, 8/1 . . . 8/31, S= 9/1 . . . 9/31, 10/1 . . . 10/31, 11/1 . . . 11/30, 12/1 . . . 12/31 . (1) The occasion H identified by the event of a July date of birth is defined by complying with 31 sample points. H = 7/1, 7/2, . . . , 7/31 . 6 (2) 3. If we need to inspect whether each resistance no fall listed below a minimum worth (in this instance 50 ohms for R1 and 100 ohms because that R2 ), a partition is C1 , C2 , C3 , C4 where C1 = {R1 C2 = {R1 (3) (4) 4. If we desire to inspect whether the resistors in parallel space within an acceptable selection of 90 to 110 ohms, a partition is D1 = (1/R1 + 1/R2 )−1 trouble 1.3.1 solution (a) A and B mutually exclusive and collectively exhaustive suggest P + P = 1. Due to the fact that P = 3 P, we have actually P = 1/4. (b) because P = P, we watch that B ⊆ A. This means P = P. Because P = 0, then P = 0. (c) since it’s always true the P = P + P − P + p
= 0.6. (1) The probability a quick call will have no handoffs is ns
= P
+ P
= 0.2. The probability the a call with one handoff will be lengthy is p
= = . P
0.2 2 (3) (c) The probability a contact is lengthy is P
ns
+ ns
0.1 + 0.2 3 = = = . Ns
= (4) difficulty 1.4.2 equipment Let si represent the outcome the the roll is i. So, because that 1 ≤ i ≤ 6, Ri = si . Similarly, Gj = sj+1 , . . . , s6 . (a) due to the fact that G1 = s2 , s3 , s4 , s5 , s6 and all outcomes have probability 1/6, P