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605-655 Level|   Probability|      
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In a certain game, you pick a card from a standard deck of 20 Dec 2012, 19:10
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72% (02:18) correct 28% (02:28) wrong based on 532 sessions

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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

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For a full discussion of this question and of "at least" probability question in general, see:
https://magoosh.com/gmat/2012/gmat-math ... -question/
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In a certain game, you pick a card from a standard deck of 28 Jun 2015, 23:25
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sunita123
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
Show more

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) = 9/16

Answer Option B
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In a certain game, you pick a card from a standard deck of 28 Jun 2015, 11:41
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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.

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B

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In a certain game, you pick a card from a standard deck of 20 Dec 2012, 22:22
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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

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 10 Jul 2016, 07:34
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mikemcgarry
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:
https://magoosh.com/gmat/2012/gmat-math- ... -question/
Show more

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
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In a certain game, you pick a card from a standard deck of 10 Jul 2016, 12:01
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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
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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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In a certain game, you pick a card from a standard deck of 24 Apr 2018, 04:04
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EMPOWERgmatRichC niks18 Bunuel amanvermagmat chetan2u

Quote:
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.
Show more

Can I also say, out of 42 cards, 13 are hearts so the probability of NOT choosing a heart is
52-13 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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In a certain game, you pick a card from a standard deck of 24 Apr 2018, 09:48
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EMPOWERgmatRichC niks18 Bunuel amanvermagmat chetan2u

Quote:
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
52-13 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
Show more

Hi adkikani

Generally 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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In a certain game, you pick a card from a standard deck of 24 Apr 2018, 15:19
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adkikani
EMPOWERgmatRichC niks18 Bunuel amanvermagmat chetan2u

Quote:
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
52-13 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
Show more

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).

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In a certain game, you pick a card from a standard deck of 17 May 2018, 13:18
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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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In a certain game, you pick a card from a standard deck of 17 May 2018, 19:36
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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.

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In a certain game, you pick a card from a standard deck of 27 Jun 2020, 11:06
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If the question were to ask what is the probability that one would win on the third draw, would the correct answer be 7/16?
1-(3/4)^2=7/16

However, since the question is asking what are the odds of having a minimum of 3 draws to gain a victory (we don't care about the the third draw), we just use (3/4)^2=9/16, correct?
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In a certain game, you pick a card from a standard deck of 28 Jun 2020, 01:54
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bones2020
If the question were to ask what is the probability that one would win on the third draw, would the correct answer be 7/16?
1-(3/4)^2=7/16

However, since the question is asking what are the odds of having a minimum of 3 draws to gain a victory (we don't care about the the third draw), we just use (3/4)^2=9/16, correct?
Show more

In order to win on precisely the third draw, three things need to happen: you must lose on the first draw AND lose on the second draw AND win on the third draw. So we need to multiply the probability of those three events together: (3/4)(3/4)(1/4) = 9/64

If we want to know the probability the game lasts at least three draws, then we just need to lose on the first two draws, so the probability of that is (3/4)(3/4) = 9/16.

If we subtract that probability from 1, to get 7/16, we're finding the probability the game does NOT last at least three draws, so 7/16 is the probability the player wins on either the first or second draw. If that were the question, you could also solve in another way, by dividing the problem into two cases: the player wins 1/4 of the time on the first draw. 3/4 of the time the player loses on the first draw, but then wins on the second draw with a probability of 1/4, so the probability of winning precisely on the second draw is (3/4)(1/4) = 3/16. Adding the probabilities of winning on the first and second draws, we get 1/4 + 3/16 = 7/16

I'd add that the question in this thread is not a realistic GMAT problem (the math involved is realistic, but the setup is not). You don't need to know anything at all about decks of cards for the test. The wording is also a lot more complicated than what you'd see on the real test.
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In a certain game, you pick a card from a standard deck of 17 Oct 2024, 11:02
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akadiyan
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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How did you get second probability of selecting heart ? 3/4*1/4 part??
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In a certain game, you pick a card from a standard deck of 17 Oct 2024, 12:34
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akadiyan
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
How did you get second probability of selecting heart ? 3/4*1/4 part??
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The question essentially asks for the probability that the game continues for at least 3 picks. The complementary event would be the game ending after 1 or 2 picks.

The game ends after the first pick if the player picks a heart right away, with a probability of 13/52 = 1/4.

The game ends after the second pick if the player picks something other than a heart on the first draw, followed by a heart on the second. The probability of this happening is 39/52 * 13/52 = 3/4 * 1/4 = 3/16.

So, the probability that the game ends after 1 or 2 picks (i.e., less than 3 picks) is 1/4 + 3/16 = 7/16.

Therefore, the probability that the game lasts at least 3 picks is P(≥ 3) = 1 - P(< 3) = 1 - 7/16 = 9/16.

Answer: B.

Hope it's clear.
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