$16$ cards. What is the conditional probability, that initially John took this card of boy?

Question

John and Jane are taking the cards from the well mixed pack of $16$ cards. In pack of cards are $4$ aces (A), $4$ kings (K), $4$ queens (Q) and $4$ boys (J). First John take one card from the top of the pack. But if he take a boy then he quickly, before Jane see, take another card and return card with boy on the top of the pack. Otherwise he keep the card.

Question Let say that Jane have a boy. What is the conditional probability, that initially John took this card of boy? My solution is $3/28$. Is that correct?

Answer 1

As calculated in the answers to your other question (which, by the way, it would be good to link to in such a case of related questions), the probability for Jane to take a J is $\frac9{20}$. The probability for Jane to take a J that John took is simply the probability that John took a J, i.e. $\frac14$. Thus the conditional probability is

$$ \frac{\frac14}{\frac9{20}}=\frac59\;. $$

Answer 2

The probability that John has taken a boy and then Jane takes that boy is $\frac{4}{16}=\frac14$

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The probability that John has taken a non-boy and then Jane takes a boy is $\frac{12}{16}\cdot\frac{4}{15}=\frac15$

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Hence the probability that Jane takes the boy that John has taken is $\frac{\frac14}{\frac14+\frac15}=\frac59$