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:

Hey Guys, i am stuck understanding the word play in these two questions:

1. Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?

A. 5/7 B. 23/24 C. 6/7 D. 34/35 E. 8/13

2. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28 B. 1/4 C. 9/16 D. 1/32 E. 11/32

Can someone help me understand how are these two questions different in terms of what has been asked.

What is the difference b/w "all not blue" and "both aren't blue out of two" ?

And why the logic fits for the 1st question and not the 2nd one. P(no blue)=1- (all blue)

Answer to 1st question can be calculated in both ways: Method 1: (3C1*4C2)/7C3 + (3C2*4C1)/7C3 + 4C3/7C3 = 18/35 + 12/35 + 4/35 = 34/35. Method 2: 1- 3C3/7C3 = 34/35

But for 2nd question: we can't use the 1 - P method: Method 1: (First card is not blue and second is not blue) 6/8 * 5/7 = 15/28 Method 2: 1-2C2/8C2 = 27/28 (And this is incorrect)
_________________

And many strokes, though with a little axe, hew down and fell the hardest-timbered oak. - William Shakespeare

So the first question is asking what is the probability that all of the beads will not be blue. That means that means that as long as all three are not blue, you are good to go. In this case the P(no blue)=1- (all blue) works: All Blue = (3/7) x (2/6) x (1/5) = 1/35 1 - (1/35) = 34/35 Answer is D

However for the second question, that is not what it is asking. It is asking for the probability that you do not get any blues. period. Here is how you solve this question (exactly what you said): (6/8) x (5/7) = 15/28 This makes sense because you it is the probability that the first card is not blue and the second card is not blue. Here is how you would solve it using the "1-x" method: probability of first card blue and second card anything else: (2/8) x (6/7) = 6/28 probability of first card anything else and second card blue: (6/8) x (2/7) = 6/28 probability that both cards are blue: (2/8) x (1/7) = 1/28 add all three together you get 13/28 1 - (13/28) = 15/28 Answer is D

So the first question is asking what is the probability that all of the beads will not be blue. That means that means that as long as all three are not blue, you are good to go. In this case the P(no blue)=1- (all blue) works: All Blue = (3/7) x (2/6) x (1/5) = 1/35 1 - (1/35) = 34/35 Answer is D

However for the second question, that is not what it is asking. It is asking for the probability that you do not get any blues. period. Here is how you solve this question (exactly what you said): (6/8) x (5/7) = 15/28 This makes sense because you it is the probability that the first card is not blue and the second card is not blue. Here is how you would solve it using the "1-x" method: probability of first card blue and second card anything else: (2/8) x (6/7) = 6/28 probability of first card anything else and second card blue: (6/8) x (2/7) = 6/28 probability that both cards are blue: (2/8) x (1/7) = 1/28 add all three together you get 13/28 1 - (13/28) = 15/28 Answer is D

Hope that helps!

Thanks, that did help. Basically the question 1 asks that all beads together be not blue. but there can be mixture of blue and non-blues, but in the second question, we don't want any blue, neither in the first place nor in the 2nd. Correct ?
_________________

And many strokes, though with a little axe, hew down and fell the hardest-timbered oak. - William Shakespeare

Hey Guys, i am stuck understanding the word play in these two questions:

1. Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?

A. 5/7 B. 23/24 C. 6/7 D. 34/35 E. 8/13

2. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28 B. 1/4 C. 9/16 D. 1/32 E. 11/32

Can someone help me understand how are these two questions different in terms of what has been asked.

What is the difference b/w "all not blue" and "both aren't blue out of two" ?

And why the logic fits for the 1st question and not the 2nd one. P(no blue)=1- (all blue)

Answer to 1st question can be calculated in both ways: Method 1: (3C1*4C2)/7C3 + (3C2*4C1)/7C3 + 4C3/7C3 = 18/35 + 12/35 + 4/35 = 34/35. Method 2: 1- 3C3/7C3 = 34/35

But for 2nd question: we can't use the 1 - P method: Method 1: (First card is not blue and second is not blue) 6/8 * 5/7 = 15/28 Method 2: 1-2C2/8C2 = 27/28 (And this is incorrect)

Hey Guys, i am stuck understanding the word play in these two questions:

1. Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?

A. 5/7 B. 23/24 C. 6/7 D. 34/35 E. 8/13

2. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28 B. 1/4 C. 9/16 D. 1/32 E. 11/32

Can someone help me understand how are these two questions different in terms of what has been asked.

What is the difference b/w "all not blue" and "both aren't blue out of two" ?

And why the logic fits for the 1st question and not the 2nd one. P(no blue)=1- (all blue)

Answer to 1st question can be calculated in both ways: Method 1: (3C1*4C2)/7C3 + (3C2*4C1)/7C3 + 4C3/7C3 = 18/35 + 12/35 + 4/35 = 34/35. Method 2: 1- 3C3/7C3 = 34/35

But for 2nd question: we can't use the 1 - P method: Method 1: (First card is not blue and second is not blue) 6/8 * 5/7 = 15/28 Method 2: 1-2C2/8C2 = 27/28 (And this is incorrect)

Thank you Bunuel, for pointing me to right post, there were nice explanations for the same concern.
_________________

And many strokes, though with a little axe, hew down and fell the hardest-timbered oak. - William Shakespeare

So the first question is asking what is the probability that all of the beads will not be blue. That means that means that as long as all three are not blue, you are good to go. In this case the P(no blue)=1- (all blue) works: All Blue = (3/7) x (2/6) x (1/5) = 1/35 1 - (1/35) = 34/35 Answer is D

However for the second question, that is not what it is asking. It is asking for the probability that you do not get any blues. period. Here is how you solve this question (exactly what you said): (6/8) x (5/7) = 15/28 This makes sense because you it is the probability that the first card is not blue and the second card is not blue. Here is how you would solve it using the "1-x" method: probability of first card blue and second card anything else: (2/8) x (6/7) = 6/28 probability of first card anything else and second card blue: (6/8) x (2/7) = 6/28 probability that both cards are blue: (2/8) x (1/7) = 1/28 add all three together you get 13/28 1 - (13/28) = 15/28 Answer is D

Hope that helps!

Thanks, that did help. Basically the question 1 asks that all beads together be not blue. but there can be mixture of blue and non-blues, but in the second question, we don't want any blue, neither in the first place nor in the 2nd. Correct ?

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...