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:

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Total outcomes = 52P13
Possible outcomes= 4.51P12 (13th card can be an ace in 4 ways)
probability = 4/52=1/13

Akamaibrah- If cards had sets of 3 Aces and not 4, then the answer would have been 3/52, therefore I am not too sure if we can solve this problem in the method you suggested. Please clarify if I am wrong.

AkamaiBrah wrote:

Bhai wrote:

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

Akmai,

I hope you can help me out with the question I have. Wouldn't this be the solution only if we assume that the each of the first 12 cards is put back in the deck?

If we assume that the cards are not put back in the deck , then I think we need to consider the followig probabilities:

(1) Probability that there is EXACTLY one Ace in the first 12 cards and the 13th card is Ace again = [12C1*(4/52)(48/52)^11] *3/39

(2) Probability that there are EXACTLY 2 aces in the first 12 cards and the 13th card is Ace again = [12C2 * (4/52)^2 * (48/52)^10] * 2/39

Similarly, the probability with (exactly 3 and 13th card ace) and with (exactly 4 and the 13 card ace) can be found.

The required probability would be the total of all the probability found above.

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

Akmai,

I hope you can help me out with the question I have. Wouldn't this be the solution only if we assume that the each of the first 12 cards is put back in the deck?

If we assume that the cards are not put back in the deck , then I think we need to consider the followig probabilities:

(1) Probability that there is EXACTLY one Ace in the first 12 cards and the 13th card is Ace again = [12C1*(4/52)(48/52)^11] *3/39

(2) Probability that there are EXACTLY 2 aces in the first 12 cards and the 13th card is Ace again = [12C2 * (4/52)^2 * (48/52)^10] * 2/39

Similarly, the probability with (exactly 3 and 13th card ace) and with (exactly 4 and the 13 card ace) can be found.

The required probability would be the total of all the probability found above.

Please clarify why this approach would be wrong.

If you add up all of the probabilities of ALL of the contigencies, the sum total will be 1/13.

Think of this: Suppose you deal 1 card off the deck. The probabilty is clearly 1/13. Now reshuffle and deal 13 cards off the top of the deck and put the cards back on the deck with the 13th card on top. Can you think of any reason why this card should have any probability other than 1/13?
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

Akmai,

I hope you can help me out with the question I have. Wouldn't this be the solution only if we assume that the each of the first 12 cards is put back in the deck?

If we assume that the cards are not put back in the deck , then I think we need to consider the followig probabilities:

(1) Probability that there is EXACTLY one Ace in the first 12 cards and the 13th card is Ace again = [12C1*(4/52)(48/52)^11] *3/39

(2) Probability that there are EXACTLY 2 aces in the first 12 cards and the 13th card is Ace again = [12C2 * (4/52)^2 * (48/52)^10] * 2/39

Similarly, the probability with (exactly 3 and 13th card ace) and with (exactly 4 and the 13 card ace) can be found.

The required probability would be the total of all the probability found above.

Please clarify why this approach would be wrong.

If you add up all of the probabilities of ALL of the contigencies, the sum total will be 1/13.

Think of this: Suppose you deal 1 card off the deck. The probabilty is clearly 1/13. Now reshuffle and deal 13 cards off the top of the deck and put the cards back on the deck with the 13th card on top. Can you think of any reason why this card should have any probability other than 1/13?

AkamaiBrah,

Thanks for the explanation. It now makes sense. It is amazing to see that the probability of all the alternatives will add up to 1/13. But I agree that in this example, it is important to look at the problem in the way you did.

gmatclubot

Re: Bhai's Probability 10
[#permalink]
03 Feb 2004, 07:55