Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

68% (01:40) correct
32% (01:21) wrong based on 53 sessions

HideShow timer Statistics

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?

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

Show Tags

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.
_________________

The only ability the GMAT is an indicator of...is the ability to do well on the GMAT.

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

Show Tags

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?

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.

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.

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

Show Tags

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

gmatclubot

Re: In a certain game, you pick a card from a standard deck of
[#permalink]
12 Jul 2016, 18:02

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...