Chip 12 going to the second player

Question

A bag holds $15$ chips labelled $1$ to $15$ . Players $A$ , $B$ and $C$ , in that order, each draw $5$ chips without replacement. You learn that every chip $A$ drew is even-numbered. Given this, what is the probability that chip $12$ ends up with $B$ ?

Short answer

$1/7$ . $A$ 's five chips are a random $5$ of the $7$ even chips, so chip $12$ is left in the bag with probability $\tfrac27$ . Then $B$ takes half of the remaining $10$ chips, so $B$ gets it with probability $\tfrac12$ ; overall $\tfrac17$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Conditional Probability, Total Probability, and Bayes' Rule
Counting and Combinatorics
Worked Examples — Probability Trees, Monty Hall, and Bayesian Updating

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