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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

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]

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11 Sep 2012, 07:08

1

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

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12 Sep 2012, 03:26

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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.

Hence Answer E.
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Re: Terry holds 12 cards, each of which is red, white, green, or [#permalink]

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12 Sep 2012, 12:38

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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

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

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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?
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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]

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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.

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

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02 Oct 2015, 00:45

I guess my approach was a little different.

S1) P(B) ~33% , that leaves roughly 67% for the rest of the probabilities. We're not given any information about R,G,W, so NS. If P(B)>50%, than this would have been sufficient. Since we don't need any information on the number or ratio of the cards to total, to answer if it's less the P(R or W)<50%. Not Sufficient

S2) P(R) ~ 17%. Same logic as S1 Applies, we can have a case where W makes up the majority of the cards, or has minimal representation. One leading to P(R or W)>50%, and the other P(R or W)<50%. Not Sufficient

S1 + S2: Known probability is 50%, leaving the other half to be shared by G or W. We can have a case where W is around 40%, leaving 10% for G, this would put P(R or W)>50%, or we can have a case where W is 10%, and G is 40%, which would mean P(R or W)<50%. Not Sufficient

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

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05 Aug 2016, 06:22

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

Practice Questions Question: 39 Page: 278 Difficulty: 650

From stimulus Total cards = 12

(1) The probability that the person will select a blue card is 1/3 Blue = 1/3 of 12 Therefore there are 4 blue cards , We dont know anything about Red, white or green INSUFFICIENT

(2) The probability that the person will select a red card is 1/6 Red = 1/6 of 12 Therefore there are 2 red cards We don't know anyting about blue green and white cards INSUFFICIENT

MERGE BOTH STATEMENTS Blue = 4 ; Red = 2 ; Red and Blue = 6 ; White and Green=6 We still don't know how many WHITE cards are there. There can be 1 or 2 or 3 or 4 or 5 white cards Therefore our probability with keep changing. P(Red)*P(White)= 1/2 * {1/12} or {2/12} or {3/12} or {4/12} or {5/12} NOT SUFFICIENT

ANSWER IS E
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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

We are given that Terry has 12 total cards. The cards are colored red, white, green, or blue. We need to determine whether the probability of selecting a red or a white card is less than ½. Remember, since we are determining the probability of selecting a red or a white card, we must add the probabilities.

Is P(red card) + P(white card) < ½?

Since the sum of all probabilities in a sample set is equal to 1, we also know that:

Thus, if we can determine the sum of the probabilities of selecting a red card and of selecting a white card OR the sum of the probabilities of selecting a blue card and of selecting a green card, we also could determine the probability of selecting a red or white card.

Statement One Alone:

The probability that the person will select a blue card is 1/3.

Since we don’t know the probability of selecting a green card, we cannot determine:

1 – [P(blue card) + P(green card)] OR

P(red card) + P(white card)

Thus, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

The probability that the person will select a red card is 1/6.

Since we don’t know the probability of selecting a white card, we cannot determine:

P(red card) + P(white card)

Thus, statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using both statements together we know the following:

P(red card) = 1/6

P(blue card) = 1/3

Substituting this into our two expressions we have:

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

Given: 12 cards - each card is red, white, green, or blue

Target question:Is the probability less than 1/2 that the card selected will be either red or white? This is a good candidate for rephrasing the target question. In order for P(selected card is red or white) < 1/2, it must be the case that there are fewer than 6 cards that are either red or white. Let R = # of red cards in the deck Let W = # of white cards in the deck Let G = # of green cards in the deck Let B = # of blue cards in the deck REPHRASED target question:Is R + W < 6?

Statement 1: The probability that the person will select a blue card is 1/3 This tells us that B = 4 (since 4/12 = 1/3) There are several CONFLICTING scenarios that satisfy statement 1. Here are two: Case a: R = 2, W = 1, G = 5 and B = 4. In this case, R + W = 2 + 1 = 3. So, R + W < 6 Case b: R = 2, W = 6, G = 0 and B = 4. In this case, R + W = 2 + 6 = 8. So, R + W > 6 Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The probability that the person will select a red card is 1/6 This tells us that R = 2 (since 2/12 = 1/6) There are several CONFLICTING scenarios that satisfy statement 2. Here are two: Case a: R = 2, W = 1, G = 5 and B = 4. In this case, R + W = 2 + 1 = 3. So, R + W < 6 Case b: R = 2, W = 6, G = 0 and B = 4. In this case, R + W = 2 + 6 = 8. So, R + W > 6 Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED. Since we cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT