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Suppose a Box Contains 6 Defective Transistors and 13 Good

Question 109

Multiple Choice

Suppose a box contains 6 defective transistors and 13 good transistors. If 2 transistors are drawn from the box without replacement, what is the probability that one of the transistors drawn is good G and one of them is defective D? ​


A) Pr(one G ∩ one D) = Suppose a box contains 6 defective transistors and 13 good transistors. If 2 transistors are drawn from the box without replacement, what is the probability that one of the transistors drawn is good G and one of them is defective D? ​ A) Pr(one G ∩ one D) = B) Pr(one G ∩ one D) = C) Pr(one G ∩ one D) = D) Pr(one G ∩ one D) = E) Pr(one G ∩ one D) =
B) Pr(one G ∩ one D) = Suppose a box contains 6 defective transistors and 13 good transistors. If 2 transistors are drawn from the box without replacement, what is the probability that one of the transistors drawn is good G and one of them is defective D? ​ A) Pr(one G ∩ one D) = B) Pr(one G ∩ one D) = C) Pr(one G ∩ one D) = D) Pr(one G ∩ one D) = E) Pr(one G ∩ one D) =
C) Pr(one G ∩ one D) = Suppose a box contains 6 defective transistors and 13 good transistors. If 2 transistors are drawn from the box without replacement, what is the probability that one of the transistors drawn is good G and one of them is defective D? ​ A) Pr(one G ∩ one D) = B) Pr(one G ∩ one D) = C) Pr(one G ∩ one D) = D) Pr(one G ∩ one D) = E) Pr(one G ∩ one D) =
D) Pr(one G ∩ one D) = Suppose a box contains 6 defective transistors and 13 good transistors. If 2 transistors are drawn from the box without replacement, what is the probability that one of the transistors drawn is good G and one of them is defective D? ​ A) Pr(one G ∩ one D) = B) Pr(one G ∩ one D) = C) Pr(one G ∩ one D) = D) Pr(one G ∩ one D) = E) Pr(one G ∩ one D) =
E) Pr(one G ∩ one D) = Suppose a box contains 6 defective transistors and 13 good transistors. If 2 transistors are drawn from the box without replacement, what is the probability that one of the transistors drawn is good G and one of them is defective D? ​ A) Pr(one G ∩ one D) = B) Pr(one G ∩ one D) = C) Pr(one G ∩ one D) = D) Pr(one G ∩ one D) = E) Pr(one G ∩ one D) =

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