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In a certain game, you pick a card from a standard deck of

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In a certain game, you pick a card from a standard deck of [#permalink]

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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 did it take before the person picked a heart and won. What is the probability that there will be at least three draws involved in a win, i.e. someone picking her first heart on the third draw or later?

(A) 1/2
(B) 9/16
(C) 11/16
(D) 13/16
(E) 15/16


For a full discussion of this question and of "at least" probability question in general, see:
http://magoosh.com/gmat/2012/gmat-math- ... -question/
[Reveal] Spoiler: OA

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Re: In a certain game, you pick a card [#permalink]

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New post 20 Dec 2012, 22:22
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 did it take before the person picked a heart and won. What is the probability that there will be at least three draws involved in a win, i.e. someone picking her first heart on the third draw or later?
(A) 1/2
(B) 9/16
(C) 11/16
(D) 13/16
(E) 15/16


Probability of picking a heart on any draw = 1/4
Probability of NOT picking a heart on the first draw AND on the second draw = [1-(1/4)] X [1-(1/4)] = 3/4 X 3/4 = 9/16

Thanks.
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In a certain game, you pick a card from a standard deck of [#permalink]

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New post 10 Jul 2016, 07:34
mikemcgarry wrote:
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 did it take before the person picked a heart and won. What is the probability that there will be at least three draws involved in a win, i.e. someone picking her first heart on the third draw or later?

(A) 1/2
(B) 9/16
(C) 11/16
(D) 13/16
(E) 15/16


For a full discussion of this question and of "at least" probability question in general, see:
http://magoosh.com/gmat/2012/gmat-math- ... -question/


Hi Mike,

Could you please help where I am going wrong. Why haven't we multiplied by 1/4?

I did this question in the mentioned way:

The probability of not drawing a heart in the first draw : 3/4
The probability of not drawing a heart in the second draw : 3/4
The probability of drawing a heart in the third draw : 3/4

So probability of not drawing a heart in first and second draw but drawing a heart in the third draw is 3/4*3/4*1/4=9/64.


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In a certain game, you pick a card from a standard deck of [#permalink]

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aks456 wrote:
Hi Mike,

Could you please help where I am going wrong. Why haven't we multiplied by 1/4?

I did this question in the mentioned way:

The probability of not drawing a heart in the first draw : 3/4
The probability of not drawing a heart in the second draw : 3/4
The probability of drawing a heart in the third draw : 3/4

So probability of not drawing a heart in first and second draw but drawing a heart in the third draw is 3/4*3/4*1/4=9/64.

Thanks

Dear aks456,
I'm happy to respond. :-)

Think about the scenario about which the question asks. The question is asking about the scenario of winning the game on the third draw or later. In other words, we know there is no Heart on the first & second draws, then anything can happen after that. That's why it's (3/4)(3/4) = 9/16.

What you did was to assume a positive result, the drawing of Heart, on exactly the third draw. The number you calculated, 9/64, is probability that the first draw of a heart occurs on exactly the third throw. This the correct calculation for another question, but not the number for which this question is asking. It's always a tricky think about probability: you have to be sure that the calculation you are doing is answering exactly the question asked, and not another question.

Does all this make sense?

Mike :-)
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Re: In a certain game, you pick a card from a standard deck of [#permalink]

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New post 12 Jul 2016, 18:02
mikemcgarry wrote:
aks456 wrote:
Hi Mike,

Could you please help where I am going wrong. Why haven't we multiplied by 1/4?

I did this question in the mentioned way:

The probability of not drawing a heart in the first draw : 3/4
The probability of not drawing a heart in the second draw : 3/4
The probability of drawing a heart in the third draw : 3/4

So probability of not drawing a heart in first and second draw but drawing a heart in the third draw is 3/4*3/4*1/4=9/64.

Thanks

Dear aks456,
I'm happy to respond. :-)

Think about the scenario about which the question asks. The question is asking about the scenario of winning the game on the third draw or later. In other words, we know there is no Heart on the first & second draws, then anything can happen after that. That's why it's (3/4)(3/4) = 9/16.

What you did was to assume a positive result, the drawing of Heart, on exactly the third draw. The number you calculated, 9/64, is probability that the first draw of a heart occurs on exactly the third throw. This the correct calculation for another question, but not the number for which this question is asking. It's always a tricky think about probability: you have to be sure that the calculation you are doing is answering exactly the question asked, and not another question.

Does all this make sense?

Mike :-)


Got it. Thanks Mike :)
Re: In a certain game, you pick a card from a standard deck of   [#permalink] 12 Jul 2016, 18:02
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