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Probability Question - Word Play [#permalink]
05 Jun 2013, 13:27

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

Re: Probability Question - Word Play [#permalink]
05 Jun 2013, 13:50

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

Re: Probability Question - Word Play [#permalink]
05 Jun 2013, 20:56

lchen wrote:

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

Re: Probability Question - Word Play [#permalink]
06 Jun 2013, 00:34

Expert's post

nikhiil wrote:

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)

Re: Probability Question - Word Play [#permalink]
06 Jun 2013, 06:36

Bunuel wrote:

nikhiil wrote:

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

Re: Probability Question - Word Play [#permalink]
06 Jun 2013, 15:18

1

This post received KUDOS

nikhiil wrote:

lchen wrote:

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 ?

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