Question

Please be as clear as possible.

Textbook - Applied Statistics and Probability for Engineers by Montgomery, 6th Edition

In a hybrid truss structure, made of steel and wood elements, we want to check their stress state. Elements can be either in compression or in tension. By observing the truss, one finds that: • 65% of the elements are in compression. • 30% of the elements in compression are in steel. • 60% of the elements in tension are in steel. An element is selected at random. a) Find the probability that: с : the element is in compression. т : the element is in tension. SIC : the element, being in compression, is in steel. ST: the element, being in tension, is in steel. b) What is the probability that a randomly picked element is in steel? c) Given that an element is in steel, what is the probability that it is in tension? Some elements are also marked (M) with a quality stamp. • 25% of all elements are marked. • 90% of the steel elements in compression are marked. An element is selected at random. d) Find the probability that: M : the element is marked. MSC : the element, being in steel and in compression, is marked. MOSC : the element is marked, in steel and in compression. (Hint: try using the multiplication rule twice; this is the so-called "chain rule") e) Find the probability that the element is in steel and in compression, given that it is marked. Express in set notation and show on Venn diagrams the following events: f) An element in compression is marked, but it is not in steel. g) An element is neither marked and in compression, nor in steel and in tension.

Answer 1

are : Given there elements either in compression or tension " the elements are in compression 30% of the elements in compression are in steel • 609. the elements in tension are in steel 65% of of 0.65 1-0.65 0.35 elements in compression / | P[c] = 65% (5) P(T) = 1- P[C] : P[SI] = {30% of of P [5/6] 65% the cleanento) / 155% of the elements} = 0- 30 x 0.65 0.195 0.30 6.65 0.65 P[SIT) = 60% of (100--05)) 60% 7 all elements in Tension 35% of all elements Elements in Tension 0.60 x 0.35 0.35 P[SIT - 0.60 0:21 0.35 6 P[s] = ? P[s] = PC SMC ) + PUSAT) = 0.30*65) + (0.60*0:25) 0.195 +0.21 P[s] 0.405 ® P[TS] =P[Element is in tension given that it is in steed] PEST] x P [T) ( Bayes The oren) P[5/7] xP (1) + PIS/CI+PCC)

P[TIS 10.60x 0.35) (0.60% 0:35) + (0 : 30 x 0'65) 0:21 0.21 0.5185 0.405 0.21 +0.195 0.5185 required probability Also Given : 25% of all elements are marked ession are the steel elements in compression 90% of marked 0.25 @ [P(M) 25% P(m/snc) = P[M Masno pesno) 6.30 x 0.65 0.195 Now, P[sno] = P[SIC] XP (C) -0.90 x 0.195 90% x 0:1959 - * P[mos no 0:1755 P[m/sno] = 0.1755 0.90 0:195 [mnsne] - 0.90 x 0.195 0.1755

P [element is in steel and in compression, ] given it is marked » Pł Sne/m] P[m/sne] « P[sn] P{M /sne] x p (s ne] + P[M/6n (7) * P[[59 ()'] 0 90 x 0.195 6.90x0.195) + (-0,90) * (1-0.195) (0.90 x 0.195) (0.90x0.195) + (0.10% 0.205) 0,1755 0.1755 = 0-6866 こ 0.256 0.1755 + 0.080s 0.6666 P[sncm