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2025 PEA Q10

Under the null p=1/2p = 1/2, compare P(15 of 20 smokers die first)P(\text{15 of 20 smokers die first}) vs.\ P(12 of 20)P(\text{12 of 20}).

Reveal answer and solution

Answer

B

Solution

  1. 1

    Each is a Binomial(20,1/2)\mathrm{Binomial}(20,1/2) probability. The ratio is

  2. 2
    P(X=15)P(X=12)=(2015)(2012). \frac{P(X=15)}{P(X=12)} = \frac{\binom{20}{15}}{\binom{20}{12}}.
  3. 3

    Now (2015)=(205)=15504\binom{20}{15} = \binom{20}{5} = 15504 and (2012)=(208)=125970\binom{20}{12} = \binom{20}{8} = 125970. Therefore

  4. 4
    15504125970≈0.123<1. \frac{15504}{125970} \approx 0.123 < 1.

Answer structure / marking notes

The mode of Bin(20,1/2)\mathrm{Bin}(20,1/2) is at 1010; outcomes farther from the mean are less likely. Since 1515 is farther from 1010 than 1212 is, the first probability is smaller.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.