Problem-Solving and Data Analysis practice question

Answer: D. 14/23

Machine A made 0.60(2,500) = 1,500 chips, of which 3% — that is, 45 — failed. Machine B made 1,000 chips, of which 7% — that is, 70 — failed. The total number of failed chips is 45 + 70 = 115, so the probability that a randomly chosen failed chip came from Machine B is 70/115 = 14/23. (A) 7/100 is P(fails | from Machine B), the reverse conditional given in the stimulus, not P(from Machine B | fails). (B) 2/5 is the share of all chips made by Machine B; it ignores the given information that the chip failed. (C) 9/23 = 45/115 is P(from Machine A | fails) — the correct denominator but the wrong machine in the numerator.

Why the other answers are wrong

A. 7/100

That's the 7% given in the passage: the probability a chip fails given it came from Machine B. The question asks the reverse — given a chip failed, how likely it came from B.

B. 2/5

That's Machine B's share of all 2,500 chips, which ignores the fact that the selected chip failed inspection. You need to compare failures to failures: 70 out of 115.

C. 9/23

You used the right denominator (115 total failures) but the wrong numerator — 45 is Machine A's failures. Machine B's 70 failures go on top.

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