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A box contains 7 identical cards numbered from 0 to 6 if two cards are drawn randomly without replacement Find the probability that : $$a) $$ The two cards have even numbers $$b)$$ The first card has an odd number and the second has an even number My turn : Using the tree diagram i got $$a) = \frac{12}{42}=\frac{2}{7}$$ $$b) = \frac{12}{42}= \frac{2}{7}$$ My question : Is there other solution using the conditional probability ?

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    $\begingroup$ Both answers are correct. Your question in the last statement is not very clear $\endgroup$ – Vizag Jun 9 at 14:40
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    $\begingroup$ If you did $\frac{4}{7} \cdot \frac{3}{6}$ on the first problem, you have effectively used conditional probability since $\frac{3}{6}$ is the conditional probability of selecting a card with an even number on the second draw given that a card with an even number was selected on the first draw. $\endgroup$ – N. F. Taussig Jun 9 at 14:46
  • $\begingroup$ How did you calculate the conditional probability which is $$\frac{3}{6}$$ @N.F.Taussing $\endgroup$ – Hussien Mohamed Jun 9 at 15:45
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No. Conditional probability doesn't help here; this really is just a straightforward probability question.

Your solution is completely correct, and is the fastest way to solve the question.

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