Independent solution

How to solve this Conditional Probability question

Setup

Setup

Let d be the total number of positions. Given a total of two affected positions, exchangeability makes the affected pair uniform among all two-position subsets.

(d−22)(d2)=4051\frac{\binom{d-2}{2}}{\binom d2}=\frac{40}{51}

Model

Model

Clear denominators and retain the feasible integer root.

51(d−2)(d−3)=40d(d−1)51(d-2)(d-3)=40d(d-1)
11d2−215d+306=(d−18)(11d−17)=011d^2-215d+306=(d-18)(11d-17)=0
d=18d=18

Compute

Compute

Protecting three specified positions leaves fifteen eligible positions for the affected pair.

Pr⁡(three specified positions unaffected)=(152)(182)\Pr(\text{three specified positions unaffected})=\frac{\binom{15}{2}}{\binom{18}{2}}
=105153=3551=0.6862745098…=\frac{105}{153}=\frac{35}{51}=0.6862745098\ldots

Answer

Answer

The conditional probability rounds to 0.686.

0.686(C)\boxed{0.686\quad\text{(C)}}