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Two cards are drawn one after another (without replacement) from a well shuffled pack of $52$ playing cards. Show the probability of getting or not getting a face card by drawing a tree diagram.

My attempt enter image description here

Is this the complete solution or something is missing?

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    Apart from your notation being rather akward., I would assume that the question is asking for the probability that *at least one of the two cards drawn* is a face card., so there is still one more step left to complete as this would correspond to $P(\{F\overline{F}\}\cup\{\overline{F}F\}\cup\{FF\})$. You are also missing the calculations of the final results of the probabilities of each event. – 2017-01-21
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    I think question is not excluding the case with both face cards. – 2017-01-21
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    @KanwaljitSingh what is the purpose of that comment? Are you somehow implying that my comment is suggesting that...because it's not. – 2017-01-21
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    The confusion displayed by the previous comments shows you are not done. Clearly state the problem; then, in add'n to the diagram, bring together the relevant probabilities that add up to the answer. – 2017-01-21
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    @BruceET, what are the relevant probabilities? I am not being able to extract them. – 2017-01-21
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    @NeWtoN $P(A\cap B)=P(A)\cdot P(B\mid A)$. I.e. in your tree diagram, the probability of arriving at a specific *leaf* of the tree is the product of the probabilities to travel along each *branch* along the way. – 2017-01-21
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    Can't say which ones are relevant without knowing _exactly_ the question being asked: 'getting or not getting' seems to have been misunderstood. – 2017-01-21

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Your tree diagram looking correct. Once cross check your answer by solving question without tree diagram.

Output from tree diagram.

Case 1 -

Probability of getting at least one face card (including both are face cards) -

$\frac{12}{52} \times \frac{11}{51} + \frac{12}{52} \times \frac{40}{51} + \frac{40}{52} \times \frac{12}{51}$

Case 2 -

Probability of getting no face card -

$\frac{40}{52} \times \frac{39}{51}$

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    how do I do that? – 2017-01-21
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    Tried to sharpen English phrases a bit (before +1). Please re-edit if you think I changed your meaning. – 2017-01-21
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    @BruceET Its perfect now. Thank you. – 2017-01-21
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    @NeWtoN: First probability in your notation is $P(FF)+P(\bar F F)+P(F\bar F).$ Second is at bottom of your tree. – 2017-01-21