Terry holds 12 cards, each of which is red, white, green

[jjjinapinch]

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

Official Guide question
Answer: E

[Brent@GMATPrepNow]

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

Official Guide question
Answer: E

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?

Aside: Here's a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

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 target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent

[Scott@TargetTestPrep]

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 space is equal to 1, we also know that:

P(red card) + P(white card) + P(blue card) + P(green card) = 1

P(red card) + P(white card) = 1 - [P(blue card) + P(green card)]

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)

For instance, if there were 4 blue, 2 red, 1 white and 5 green balls, the probability of selecting white or red card is (2 + 1)/12 = 3/12 = 1/4. On the other hand, if there were 4 blue, 2 red, 4 white and 2 green balls, the probability of selecting a white or red card is (2 + 4)/12 = 6/12 = 1/2.

Thus, statement one is not sufficient to answer the question.

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)

We can take the above example of 4 blue, 2 red, 1 white and 5 green balls versus 4 blue, 2 red, 4 white and 2 green balls (notice that probability of a red card is 2/12 = 1/6 in either case), we see that we get different probabilities for a red or white card.
Thus, statement two is not sufficient to answer the question.

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:

P(red card) + P(white card) = 1/6 + P(white card) = ?

1 - [1/3 + P(green card)] = ?

We see that we still do not have enough information to determine whether the probability of selecting a red card or a white card is less than ½. In the examples above, notice that probability of selecting a blue card is 4/12 = 1/3 and probability of selecting a red card is 2/12 = 1/6. Those two examples show us that we get different probabilities of selecting a white or red card depending on the number of green and white cards in the deck.

Answer: E