GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Nov 2019, 04:00

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

James and Colleen are playing basketball. The probability of James mis

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
User avatar
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 327
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
James and Colleen are playing basketball. The probability of James mis  [#permalink]

Show Tags

New post 27 Aug 2010, 19:39
8
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

51% (01:39) correct 49% (01:59) wrong based on 59 sessions

HideShow timer Statistics

James and Colleen are playing basketball. The probability of James missing a shot is x, and the probability of Colleen not making a shot is y. If they each take 2 shots. What is the probability that they both make at least 1 shot apiece?

A) \(1 - (x^2*y^2)\)
B) \((1-x^2) (1-y^2)\)
C) \((1-(1-x)^2) (1- (1-y)^2)\)
D) \((1-(1-x)^2) ((1-y)^2)\)
E) \((1-(1-y^2)) (1-x^2)\)

I could not solve this problem and I wanted to arrive at the answer using the algebraic route and by not picking numbers.

_________________
:good Support GMAT Club by putting a GMAT Club badge on your blog :thanks
GMAT Tutor
avatar
G
Joined: 24 Jun 2008
Posts: 1827
Re: James and Colleen are playing basketball. The probability of James mis  [#permalink]

Show Tags

New post 27 Aug 2010, 20:35
1
1
ezhilkumarank wrote:
Question: James and Colleen are playing basketball. The probability of James missing a shot is x, and the probability of Colleen not making a shot is y. If they each take 2 shots. What is the probability that they both make at least 1 shot apiece?

A) \(1 - (x^2*y^2)\)
B) \((1-x^2) (1-y^2)\)
C) \((1-(1-x)^2) (1- (1-y)^2)\)
D) \((1-(1-x)^2) ((1-y)^2)\)
E) \((1-(1-y^2)) (1-x^2)\)

I could not solve this problem and I wanted to arrive at the answer using the algebraic route and by not picking numbers.


The probability James misses both shots is x*x = x^2. So the probability James makes at least one shot is 1 - x^2. Similarly, the probability Colleen makes at least one shot is 1 - y^2. To find the probability they both make at least one shot, we multiply the probability that each makes at least one shot: (1 - x^2)(1 - y^2).
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Senior Manager
Senior Manager
User avatar
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 327
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Re: James and Colleen are playing basketball. The probability of James mis  [#permalink]

Show Tags

New post 27 Aug 2010, 21:16
IanStewart wrote:
ezhilkumarank wrote:
Question: James and Colleen are playing basketball. The probability of James missing a shot is x, and the probability of Colleen not making a shot is y. If they each take 2 shots. What is the probability that they both make at least 1 shot apiece?

A) \(1 - (x^2*y^2)\)
B) \((1-x^2) (1-y^2)\)
C) \((1-(1-x)^2) (1- (1-y)^2)\)
D) \((1-(1-x)^2) ((1-y)^2)\)
E) \((1-(1-y^2)) (1-x^2)\)

I could not solve this problem and I wanted to arrive at the answer using the algebraic route and by not picking numbers.


The probability James misses both shots is x*x = x^2. So the probability James makes at least one shot is 1 - x^2. Similarly, the probability Colleen makes at least one shot is 1 - y^2. To find the probability they both make at least one shot, we multiply the probability that each makes at least one shot: (1 - x^2)(1 - y^2).



Thanks IanStewart. +1 from me.
_________________
:good Support GMAT Club by putting a GMAT Club badge on your blog :thanks
Intern
Intern
avatar
B
Joined: 01 Jun 2019
Posts: 9
Re: James and Colleen are playing basketball. The probability of James mis  [#permalink]

Show Tags

New post 20 Oct 2019, 05:13
Could someone tell me how the solution would change if the question asked for the probability that both make exactly one shot?
GMAT Club Bot
Re: James and Colleen are playing basketball. The probability of James mis   [#permalink] 20 Oct 2019, 05:13
Display posts from previous: Sort by

James and Colleen are playing basketball. The probability of James mis

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne