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A standard deck of 52 cards has 13 kinds of cards, with four cards of each kind, one in each of the four suits, hearts, diamonds, spades, and clubs. What is the probability that a five-card poker hand contains the ace of hearts?

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closed as off-topic by quid, drhab, Thomas, TravisJ, Daniel Fischer Sep 22 '15 at 13:08

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – quid, drhab, Thomas, TravisJ, Daniel Fischer
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  • $\begingroup$ Please elaborate on: 1. What is "poker hand"? 2. What have you tried? $\endgroup$ – barak manos Sep 22 '15 at 11:54
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    $\begingroup$ Hint: easier to find the probability that it does not contain the given card, and then subtract. $\endgroup$ – lulu Sep 22 '15 at 11:54
  • $\begingroup$ Hint: If $5$ cards are picked out of $52$ then what is the probability of the ace of hearts to be one of them? $\endgroup$ – drhab Sep 22 '15 at 11:54
  • $\begingroup$ What ve you tried? $\endgroup$ – Abhishekstudent Sep 22 '15 at 11:56
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Hint:

Think of it as if you are the ace of hearts yourself. If $5$ cards are taken, then what is your chance to be one of them?

If you are one of $52$ persons of wich $5$ are elected, all with equal probability. Then what is your chance to be elected?

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The probability of the hand NOT containing the ace of hearts is.... 51/52 x 50/51 x 49/50 x 48/49 x 47/48 = 0.9038 So the probability of the hand containing the ace of hearts is 1 - 0.903 = 0.0962

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  • $\begingroup$ An additional comment about the advantage of cancelling would improve you answer. $\endgroup$ – callculus Sep 22 '15 at 12:14
  • $\begingroup$ In my view cancelling is not the most effective method here. $\endgroup$ – drhab Sep 22 '15 at 12:21
  • $\begingroup$ @drhab What do you mean ? $\endgroup$ – callculus Sep 22 '15 at 12:22
  • $\begingroup$ Simply that there is a better method in this case. See my answer for that. $\endgroup$ – drhab Sep 22 '15 at 12:23
  • $\begingroup$ @drhab My comment referred to the answer of Malcolm, not to your answer. $\endgroup$ – callculus Sep 22 '15 at 12:38

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