Probability problem about a card located in a wardrobe or in $1$ of $5$ drawers each of them have the same probability

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A card is either in a wardrobe with a probability of $\frac{1}{5}$ or in one of five drawers with the same probability each drawer. If we have opened four drawers, chosen at random, and the card is not there, what is the probability that it is in the fifth drawer?

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If $W$ is the probability of the card being in the wardrobe, and $D$ is the probability of it being in the remaining drawer, we have $$P(D\mid W\cup D)=\frac{P(D\cap(D\cup W))}{P(D\cup W)}=\frac{P(D)}{P(D\cup W)}=\frac{\frac{4}{25}}{\frac{9}{25}}=\frac49$$ One notes that this is the ratio of the original probabilities of being in that particular drawer to being in that drawer or the wardrobe.