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Here's an illustration , suppose I have of deck of 52 cards(well shuffled) and I pick 10 cards at random and now again I pick a card from those 10 cards,

(I)what is the probability that it's ace of diamonds?

(II)what is the probability that a cards drawn is a face card. By deck of cards I mean normal playing cards

I feel the answer should be 1/52. I have tried searching this over web but no success.

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  • $\begingroup$ Is the second pick among the 10 cards or the remaining 42 cards? $\endgroup$
    – user_194421
    Feb 4, 2018 at 7:40
  • $\begingroup$ If you replaced "ace of diamonds" by "four of hearts" you'd expect the same answer, surely? By "queen of diamonds", by "six of hearts" etc. $\endgroup$
    – Angina Seng
    Feb 4, 2018 at 7:41
  • $\begingroup$ second pick is from 10 cards , I'm sorry for the vague question thanks for comment $\endgroup$
    – Chemist
    Feb 4, 2018 at 8:15
  • $\begingroup$ I have solved the second question have a look at it. Well I am not 100% sure of its correctness. $\endgroup$
    – dssknj
    Feb 6, 2018 at 16:40
  • $\begingroup$ can please solve the second part so that i clear my doubt completely $\endgroup$
    – Chemist
    Feb 8, 2018 at 13:53

4 Answers 4

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The probability of drawing out an ace of diamond from those 10 cards depend on whether it is in those 10 cards or not. The probability of the ace of diamond being in the first 10 cards that were drawn is ${10\over 52}$ and probability of drawing an ace of diamond out of those 10 card is ${1\over 10}$. So in total the probability of drawing an ace of diamond is $${10\over 52}*{1\over 10}={1\over 52}$$ So yes, your answer is right. The probability is not changed, its still ${1\over 52}$. I think we can use proportion for the part ii). As saying I have got 5 marks out of 10 is equivalent to I have got 50 out of 100. Similarly there are 12 face cards in 52 cards is equivalent to we have ${30\over 13}$ face cards in those 10 cards. So the probability of choosing a face cards from those 10 cards is $${{30\over 13}\over 10} = {3\over 13}$$ The first part can be solved in the same way.

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  • $\begingroup$ Your welcome. I learned from this too. $\endgroup$
    – dssknj
    Feb 4, 2018 at 17:04
  • $\begingroup$ can you help me with the second question. $\endgroup$
    – Chemist
    Feb 6, 2018 at 14:21
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It is easy to see that every card has the same probability (i.e. there is symmetry). If that probability is $p$, then the probability of picking any specific card is $\frac{p}{52p}=\frac1{52}$.

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You're right. You can reason that there are $51!/41!$ ways to choose 11 cards in order with the ace of diamonds last, out of a possible $52!/41!$ ways to pick 11 cards.

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  • $\begingroup$ I meant I'm picking a card from those 10 picked ones $\endgroup$
    – Chemist
    Feb 4, 2018 at 8:10
  • $\begingroup$ I'm sorry for the vague question, i have edited it $\endgroup$
    – Chemist
    Feb 4, 2018 at 8:12
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Its wrong . This is a classical probability question , if 10 cards are selected from 52 , the number of diamond , aces , queens must be know before you can take the probability , the no o diamonds in the 10 cards was not give, so you can't take the probability

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  • $\begingroup$ but its not known for sure about what could be there in those 10 cards , this factor makes the question difficult to answer $\endgroup$
    – Chemist
    Feb 4, 2018 at 8:47

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