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2026 PEA Q15

A restaurant has 60%60\% vegetarian customers (of whom 30%30\% order dessert) and 40%40\% non-vegetarian (of whom 50%50\% order dessert). Given that a randomly selected customer ordered dessert, what is the probability they ordered vegetarian?

Reveal answer and solution

Answer

B

Solution

  1. 1

    By Bayes' theorem

  2. 2
    P(V∣D)=P(D∣V) P(V)P(D∣V)P(V)+P(D∣Vc)P(Vc)=0.3⋅0.60.3⋅0.6+0.5⋅0.4=0.180.38=919≈0.474. P(V \mid D) = \frac{P(D \mid V)\,P(V)}{P(D \mid V) P(V) + P(D \mid V^c) P(V^c)} = \frac{0.3 \cdot 0.6}{0.3 \cdot 0.6 + 0.5 \cdot 0.4} = \frac{0.18}{0.38} = \frac{9}{19} \approx 0.474.

Answer structure / marking notes

No major trap beyond standard calculation care.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2026/ISI_MSQE_PEA_2026_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.