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There are 13 hearts in a full deck of 52 cards. In a certain game, you
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28 Jun 2015, 10:27
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There are 13 hearts in a full deck of 52 cards. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure how many draws it took before the person picked a heart and, thereby, won. What is the probability that one will pick the first heart on the third draw or later? A 1/2 B 9/16 C 11/16 D 13/16 E 15/16
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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you
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28 Jun 2015, 10:41
Hi sunita123, Here, we're told that 13 of the 52 cards are "hearts"  this is 1/4 of the deck. We're also told that after drawing a card, the card is PUT BACK in to the deck (so there will always be 52 cards to draw from). To draw a heart for the first time on the 3rd draw (or later), the FIRST TWO draws MUST NOT be hearts.... Probability of NOT drawing a heart on the first draw = 3/4 Probability of NOT drawing a heart on the second draw = 3/4 (3/4)(3/4) = 9/16 From this point, you will eventually draw a heart (which is what the question is asking for), so the probability is 9/16. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you
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28 Jun 2015, 22:25
sunita123 wrote: There are 13 hearts in a full deck of 52 cards. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure how many draws it took before the person picked a heart and, thereby, won. What is the probability that one will pick the first heart on the third draw or later?
A 1/2 B 9/16 C 11/16 D 13/16 E 15/16 Favorable case = the heart is picked in the third draw or later Unfavorable case = The heart is picked in either first draw or in second draw Probability = Favorable outcomes / Total out comes
Also probability = 1(Unfavorable outcomes / Total out comes)
Unfavorable case 1: probability of heart picked in first draw = 13/52 =1/4 Unfavorable case 2: probability of heart picked in Second draw (I.e. first draw is not Heart)= (39/52)*(13/52) =(3/4)*(1/4)= 3/16 Total Unfavorable Probability = (1/4)+(3/16) = 7/16 I.e. Favorable Probability = 1  (7/16) = 9/16Answer Option B
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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you
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24 Apr 2018, 03:04
EMPOWERgmatRichC niks18 Bunuel amanvermagmat chetan2uQuote: Here, we're told that 13 of the 52 cards are "hearts"  this is 1/4 of the deck. We're also told that after drawing a card, the card is PUT BACK in to the deck (so there will always be 52 cards to draw from).
To draw a heart for the first time on the 3rd draw (or later), the FIRST TWO draws MUST NOT be hearts....
Probability of NOT drawing a heart on the first draw = 3/4 Probability of NOT drawing a heart on the second draw = 3/4
(3/4)(3/4) = 9/16
From this point, you will eventually draw a heart (which is what the question is asking for), so the probability is 9/16. Can I also say, out of 42 cards, 13 are hearts so the probability of NOT choosing a heart is 5213 ie 39 (Favorable outcomes) divided by all possible outcomes ie 52. 39/52 = 3/4 Same for the second draw: 3/4 Total Probability = 9/16
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There are 13 hearts in a full deck of 52 cards. In a certain game, you
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24 Apr 2018, 08:48
adkikani wrote: EMPOWERgmatRichC niks18 Bunuel amanvermagmat chetan2uQuote: Here, we're told that 13 of the 52 cards are "hearts"  this is 1/4 of the deck. We're also told that after drawing a card, the card is PUT BACK in to the deck (so there will always be 52 cards to draw from).
To draw a heart for the first time on the 3rd draw (or later), the FIRST TWO draws MUST NOT be hearts....
Probability of NOT drawing a heart on the first draw = 3/4 Probability of NOT drawing a heart on the second draw = 3/4
(3/4)(3/4) = 9/16
From this point, you will eventually draw a heart (which is what the question is asking for), so the probability is 9/16. Can I also say, out of 42 cards, 13 are hearts so the probability of NOT choosing a heart is 5213 ie 39 (Favorable outcomes) divided by all possible outcomes ie 52. 39/52 = 3/4 Same for the second draw: 3/4 Total Probability = 9/16 Hi adkikaniGenerally the definition of "favorable" in Probability is the event that you are looking for. So by definition 39 is your unfavorable outcome. Your calculation is perfectly ok here. First two events are unfavorable and you are multiplying those events to get the final answer.



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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you
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24 Apr 2018, 14:19
adkikani wrote: EMPOWERgmatRichC niks18 Bunuel amanvermagmat chetan2uQuote: Here, we're told that 13 of the 52 cards are "hearts"  this is 1/4 of the deck. We're also told that after drawing a card, the card is PUT BACK in to the deck (so there will always be 52 cards to draw from).
To draw a heart for the first time on the 3rd draw (or later), the FIRST TWO draws MUST NOT be hearts....
Probability of NOT drawing a heart on the first draw = 3/4 Probability of NOT drawing a heart on the second draw = 3/4
(3/4)(3/4) = 9/16
From this point, you will eventually draw a heart (which is what the question is asking for), so the probability is 9/16. Can I also say, out of 42 cards, 13 are hearts so the probability of NOT choosing a heart is 5213 ie 39 (Favorable outcomes) divided by all possible outcomes ie 52. 39/52 = 3/4 Same for the second draw: 3/4 Total Probability = 9/16 Hi adkikani, Yes  that way of writing out the math is the exact same calculation that I did (I just reduced the fraction immediately, then subtracted  whereas you subtracted first, then reduced the fraction). GMAT assassins aren't born, they're made, Rich
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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you
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17 May 2018, 12:18
Hello guys! I need help form those of you who are great at probabilities.
In this one I thought "Oh, ok. So I could calculate the complement of the probability of reaching the third trial and NOT draw a heart: 1(3/4)^3 = 37/64.
What was wrong with my train of thought?



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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you
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17 May 2018, 18:36
Hi JohnAHD, Your calculation is based on the idea that the FIRST 3 cards are NOT hearts. However, the prompt asks for the probability of pulling a heart on the 3rd card or later. GMAT assassins aren't born, they're made, Rich
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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you
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21 May 2018, 20:53
There are 13 hearts in a full deck of 52 cards. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure how many draws it took before the person picked a heart and, thereby, won. What is the probability that one will pick the first heart on the third draw or later? Probability of selecting heart in first draw = \(\frac{13}{52}\) = \(\frac{1}{4}\) Probability of selecting heard in second draw = \(\frac{3}{4}\) * \(\frac{1}{4}\) =\(\frac{3}{16}\) Probability of selecting heart in 1st and second draw = \(\frac{1}{4}\) + \(\frac{3}{16}\) = \(\frac{7}{16}\) Probability of selecting heart in 3rd draw = 1\(\frac{7}{16}\) = \(\frac{9}{16}\) Ans: B
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Re: There are 13 hearts in a full deck of 52 cards. In a certain game, you &nbs
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