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Terry holds 12 cards, each of which is red, white, green, or [#permalink]
11 Sep 2012, 03:52

Expert's post

12

This post was BOOKMARKED

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A

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E

Difficulty:

25% (medium)

Question Stats:

71% (02:12) correct
29% (01:04) wrong based on 570 sessions

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?

(1) The probability that the person will select a blue card is 1/3 (2) The probability that the person will select a red card is 1/6

Practice Questions Question: 39 Page: 278 Difficulty: 650

Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
11 Sep 2012, 03:52

Expert's post

SOLUTION

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?

The question asks whether \(\frac{red+white}{12}<\frac{1}{2}\) --> is \(red+white<6\). So, basically we need to know whether the number of red or white cards is less than 6.

(1) The probability that the person will select a blue card is 1/3 --> the number of blue cards is \(\frac{1}{3}*12=4\). Now, if there is only 1 green card then the number of red or white cards is 12-(4+1)=7 but if there are 3 green cards, then the number of red or white cards is 12-(4+3)=5. Not sufficient.

(2) The probability that the person will select a red card is 1/6 --> the number of red cards is \(\frac{1}{6}*12=2\). Not sufficient since we don't know the number of white cards.

(1)+(2) We know that there are 4 blue and 2 red cards, but we still don't know how many white cards are there: if there is only one, then the answer is YES but if there are 5 then the answer is NO. Not sufficient.

Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
11 Sep 2012, 07:08

1

This post received KUDOS

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?

(1) The probability that the person will select a blue card is 1/3 (2) The probability that the person will select a red card is 1/6

Can we determine if there are less than 6 total R or W's in the pack?

(1) There is 4 blues so 8 other cards. Therefore R+W <8, if 7 no, if 5, yes. INEFF

(2) There is 2 reds so 10 other cards. At least 1B and 1G so there is at most 8 other Whites. So at most 10/12 can be R+W, at least 3 INEFF

Together if we say number of R+W = x (i) says 2<x<8, (ii) says 1<x<10 so both together INEFF

Therefore E _________________

If you find my post helpful, please GIVE ME SOME KUDOS!

Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
12 Sep 2012, 03:26

2

This post received KUDOS

Bunuel wrote:

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?

(1) The probability that the person will select a blue card is 1/3 (2) The probability that the person will select a red card is 1/6

ST 1: Insufficient: P(blue) = 1/3, means there are 4 blue cards. So remaining are 8 cards, but don't know exact distribution. If we consider green =1, P(R+W) >1/2, but if we consider green = 4, P(R+W)<1/2. So insufficient.

ST2: Insufficient: P(Red) = 1/6, means there are 2 red cards. So remaining are 10, but don't know exact distribution. If we take white as 5 P(R+W)>1/2, If we take white 2, P(R+W)<1/2. So insufficient.

St 1 + St 2: Insufficient: Red =2, Blue = 4, Remaining 6 cards. If we take white 5, P(R+W) >1/2, But if we take White 2 P(R+W) <1/2. So insufficient.

Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
12 Sep 2012, 12:38

2

This post received KUDOS

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white? (1) The probability that the person will select a blue card is 1/3 (2) The probability that the person will select a red card is 1/6

Trick- There is no need to calculate Probability as the question ask about the total no of Red & White cards. Basically the question can be restated as is the number of Red & White cards less than 6-- Red + White <6

Red+ White + Green + Blue = 12 Statement 1 - Probability of Blue = 1/3 = 4 blue cards are there ---->No info is given regarding Red & White----->Insufficient Statement 2 - Probability of Red = 1/6 = 2 red cards are there ---->No info is given regarding White cards----->Insufficient Statement 1 & 2 - Red+ White + Green + Blue = 12 2+ White + Green + 4 = 12 -----> White + Green = 6 Now green can be any number from 0 to 6 i.e. Red + white can be 2,3,4,5,6,7,8----> Insufficient

Answer E

Hope it helps _________________

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Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
14 Sep 2012, 05:51

1

This post received KUDOS

Expert's post

SOLUTION

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?

The question asks whether \(\frac{red+white}{12}<\frac{1}{2}\) --> is \(red+white<6\). So, basically we need to know whether the number of red or white cards is less than 6.

(1) The probability that the person will select a blue card is 1/3 --> the number of blue cards is \(\frac{1}{3}*12=4\). Now, if there is only 1 green card then the number of red or white cards is 12-(4+1)=7 but if there are 3 green cards, then the number of red or white cards is 12-(4+3)=5. Not sufficient.

(2) The probability that the person will select a red card is 1/6 --> the number of red cards is \(\frac{1}{6}*12=2\). Not sufficient since we don't know the number of white cards.

(1)+(2) We know that there are 4 blue and 2 red cards, but we still don't know how many white cards are there: if there is only one, then the answer is YES but if there are 5 then the answer is NO. Not sufficient.

Answer: E.

Kudos points given to everyone with correct solution. Let me know if I missed someone. _________________

Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
01 Feb 2013, 23:32

Bunuel wrote:

SOLUTION

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?

The question asks whether \(\frac{red+white}{12}<\frac{1}{2}\) --> is \(red+white<6\). So, basically we need to know whether the number of red or white cards is less than 6.

(1) The probability that the person will select a blue card is 1/3 --> the number of blue cards is \(\frac{1}{3}*12=4\). Now, if there is only 1 green card then the number of red or white cards is 12-(4+1)=7 but if there are 3 green cards, then the number of red or white cards is 12-(4+3)=5. Not sufficient.

(2) The probability that the person will select a red card is 1/6 --> the number of red cards is \(\frac{1}{6}*12=2\). Not sufficient since we don't know the number of white cards.

(1)+(2) We know that there are 4 blue and 2 red cards, but we still don't know how many white cards are there: if there is only one, then the answer is YES but if there are 5 then the answer is NO. Not sufficient.

Answer: E.

but don't we know that between red and white there are 6 cards remaining so we get red + white < 6 so it will be a NO in all cases or have I missed something over here? _________________

Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
01 Feb 2013, 23:48

1

This post received KUDOS

Expert's post

fozzzy wrote:

Bunuel wrote:

SOLUTION

Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?

The question asks whether \(\frac{red+white}{12}<\frac{1}{2}\) --> is \(red+white<6\). So, basically we need to know whether the number of red or white cards is less than 6.

(1) The probability that the person will select a blue card is 1/3 --> the number of blue cards is \(\frac{1}{3}*12=4\). Now, if there is only 1 green card then the number of red or white cards is 12-(4+1)=7 but if there are 3 green cards, then the number of red or white cards is 12-(4+3)=5. Not sufficient.

(2) The probability that the person will select a red card is 1/6 --> the number of red cards is \(\frac{1}{6}*12=2\). Not sufficient since we don't know the number of white cards.

(1)+(2) We know that there are 4 blue and 2 red cards, but we still don't know how many white cards are there: if there is only one, then the answer is YES but if there are 5 then the answer is NO. Not sufficient.

Answer: E.

but don't we know that between red and white there are 6 cards remaining so we get red + white < 6 so it will be a NO in all cases or have I missed something over here?

Total = 12 _________ Blue = 4 Red = 2 White = ? Green = ?

If there is 1 white card and 5 green cards, then red+white=3<6. If there are 5 white cards and 1 green card, then red+white=6.

Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
26 Jul 2014, 01:42

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Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]
20 Jul 2015, 12:10

You will ultimately draw the same conclusion as the above methodologies, but my approach was slightly different:

Given: R + W + B + G = 12; prove that either (1) R + W <6 OR (2) B + G > 6.

(1) B = 1/3 * 12 = 4; thus, can simplify (2): (4) + G > 6? G > 2? Since we have not been provided any information pertaining to the value of G, insufficient.

(2) R = 1/6 * 12 = 2; thus, can simplify (1): (2) + W < 6? W < 4? Since we have not been provided any information pertaining to the value of W, insufficient.

Combo: Can we answer either from S1 G > 2 or from S2 W < 4? Using the information provided in both statements, (2) + W + (4) + G = 12; thus, we can deduce that W + G = 6.

W = 2, G = 3 YES W = 5, G = 1 NO

Hence, (E)

gmatclubot

Re: Terry holds 12 cards, each of which is red, white, green, or
[#permalink]
20 Jul 2015, 12:10

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