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If I have a standard deck of cards(52 cards,no jacks,4 suits). What is the probability that you will draw a prime numbered card and then draw an even numbered card.

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    $\begingroup$ When you say "no jacks", I assume you mean "no jokers" $\endgroup$ – Henry Aug 4 '18 at 23:14
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    $\begingroup$ Welcome to MSE. Please include details by editing your question, to show that you have worked on it, and explain where you are stuck. $\endgroup$ – Arnaud Mortier Aug 4 '18 at 23:15
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    $\begingroup$ Does Jack equal $11$ and therefore a prime? Does Queen equal $12$ and therefore even? Does King equal $13$ and therefore a prime? $\endgroup$ – Brian Tung Aug 4 '18 at 23:25
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Making the assumption that the draws are without replacement:

How many odd prime-numbered cards are there? Call this number $m$.

How many non-prime even-numbered cards are there? Call this number $n$.

How many cards are both prime-numbered and even-numbered? Call this number $p$.

The probability of drawing an odd prime on the first card and an even number on the second card is then $\frac{m}{52} \times \frac{\text{what}}{51}$?

The probability of drawing an even prime on the first card and an even number on the second card is then $\frac{p}{52} \times \frac{\text{what}}{51}$?

Add these last two probabilities together to obtain your answer.

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  • $\begingroup$ what is the n used for and what is the "what" in the numerator $\endgroup$ – Matthew Weppenaar Aug 5 '18 at 20:23
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Primes are: ${2,3,5,7}$ (Thus the probability of drawing one is $P_1=16/52$.)
Even numbers are: ${2,4,6,8,10}$ ($P_2=20/52$)
Assuming you are drawing with replacing it's therefore $P_1*P_2 = ~0.12$.

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  • $\begingroup$ Welcome to MSE. This should be a comment rather than an answer. Answers are supposed to be as complete as possible. Comments are there for questions and partial hints. $\endgroup$ – Arnaud Mortier Aug 4 '18 at 23:23
  • $\begingroup$ Your expression seems like it would only hold with replacement. Without replacement there are only $51$ cards left for the second draw. $\endgroup$ – Brian Tung Aug 4 '18 at 23:26
  • $\begingroup$ @ArnaudMortier math.meta.stackexchange.com/questions/21450/hints-as-answers suggests that hints are not discouraged when suitable $\endgroup$ – Henry Aug 4 '18 at 23:26
  • $\begingroup$ @ArnaudMortier: I believe this answer is suitable provided the hint is well explained. $\endgroup$ – Brian Tung Aug 4 '18 at 23:27
  • $\begingroup$ @Brian Tung: Yes, I said "with replacing" for that reason, but initially asked to clarify. I have deleted the question now to avoid confusion between the clarification and my answer. $\endgroup$ – Algebreaker Aug 4 '18 at 23:28

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